#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack> 
#include <set>
#include <map>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
using edge = struct {ll to; ll cost;};
using point = struct {ll x; ll y;};
typedef string str;
typedef std::pair<ll,ll> pl;
typedef std::pair<ll,pl> pl3;
typedef std::map<string, ll> msl;
typedef std::map<char, ll> mcl;
typedef std::map<ll, ll> mll;
typedef std::vector<ll> vl;
typedef std::vector<pl> vpl;
typedef std::vector<point> points;
typedef std::vector<pl3> vpl3;
typedef std::priority_queue<ll> pq;
typedef std::priority_queue<ll, vl, greater<ll>> pql; // priority queue taking from the lower value.
typedef std::vector<edge> Graph;
const ll N_DIGITS = 60;
const ll MOD = 1e9 + 7;
// const ll MOD = 998244353;
const ll INF = MOD * MOD;
const long double EPS = 1e-9;
const long double PI = 3.14159265358979323846;
vpl dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1},  // four directions
            {1, 1}, {-1, 1}, {-1, -1}, {1, -1},  // diagonal
            {0, 0}  // self
            };
struct undirected_edge {
    ll from; ll to; ll cost;
    bool operator<(const undirected_edge& ue) const{
        return cost < ue.cost;
    }
};

#define rep(i, n) for(ll (i) = 0; (i) < (n) ; (i)++ )
#define revrep(i, n) for(ll (i) = n - 1; (i) >= 0 ; (i)--)
#define For(i, a, b) for(ll (i) = (a); (i) < (b) ; (i)++ )
#define revFor(i, b, a) for(ll (i) = (b) - 1; (i) >= (a) ; (i)--)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define isUpper(c) ('a' - c > 0)
#define isNum(c) (0 <= (c) - '0' && (c) - '0' <= 9)
#define toLower(c) char((c)+0x20)
#define toUpper(c) char((c)-0x20)
#define pb push_back
#define mp make_pair
#define pr(a) cout << (a)
#define prl(a) cout << (a) << endl
#define prl2(a,b) cout << (a) << " " << (b) << endl
#define prl3(a,b,c) cout << (a) << " " << (b) << " " << (c) << endl
#define prl4(a,b,c,d) cout << (a) << " " << (b) << " " << (c) << " " << (d) << endl
#define prs(a) cout << (a) << " "
#define yn(condition) if ((condition)) prl("Yes"); else prl("No");
#define YN(condition) if ((condition)) prl("YES"); else prl("NO");
#define in1(a) cin >> (a)
#define in2(a,b) cin >> (a) >> (b)
#define in3(a,b,c) cin >> (a) >> (b) >> (c)
#define in4(a,b,c,d) cin >> (a) >> (b) >> (c) >> (d)
#define e1 first
#define e2 second
#define ctol(c) ll((c)) - ll('0')
#define ltos(n) to_string((n))
#define items(kv, v) for(auto &(kv): (v))
#define ndig(N, n) ctol(ll(ltos((N))[ll(ltos((N)).length()) - (n)]))
#define rsort(a,n) sort(a,a+n,greater<>())
#define Forchar(c,a,z) for(char (c) = (a); (c) <= (z); (c)++)
#define cntchar(s,c) count(all((s)), c)
#define substring(s, start, end) s.substr((start), (end) - (start) + 1)
#define prl_nd(num, digits) cout << fixed << setprecision(digits) << (num) << endl;
#define XOR(a, b) (a)^(b)
#define prl_time(s) prl3("Elapsed Time:", 1000.0 * (clock() - s) / CLOCKS_PER_SEC, "[ms]");
#define char_to_str(c) string(1, (c))
#define DEBUG(x) cerr<<#x<<": "<<(x)<<endl
#define DEBUG_VEC(v) cerr<<#v<<": ";rep(__i,(v).size())cerr<<((v)[__i])<<", ";cerr<<endl

struct MaxFlow {
    struct F_edge {
        ll to, rev, capacity;
        F_edge(ll to, ll rev, ll capacity) : to(to), rev(rev), capacity(capacity) {}
    };

    typedef vector<F_edge> F_edges;
    vector<F_edges> graph;
    ll n_vertex;
    // level is the shortest path to get a given node from the source node.
    vl level, iter;

    MaxFlow(ll n_vertex) : n_vertex(n_vertex) {graph.resize(n_vertex);}

    void add_edge(ll from, ll to, ll capacity) {
        graph[from].pb({to, ll(graph[to].size()), capacity});
        graph[to].pb({from, ll(graph[from].size()) - 1, 0});
    }

    void bfs(ll source){
        level = vl(n_vertex, -1); level[source] = 0;
        queue<ll> q; q.push(source);
        while (!q.empty()) {
            ll vertex = q.front(); q.pop();
            rep(i, graph[vertex].size()){
                ll target = graph[vertex][i].to;
                ll cap_target = graph[vertex][i].capacity;
                // if the flow can be into the target node, implement below.
                if (cap_target > 0 && level[target] < 0){
                    level[target] = level[vertex] + 1;
                    q.push(target);
                }
            }
        }
    }

    ll dfs(ll vertex, ll sink, ll flow) {
        if (vertex == sink) return flow;
        for(ll &i = iter[vertex]; i < graph[vertex].size(); i++){
            ll target = graph[vertex][i].to; ll cap_target = graph[vertex][i].capacity;
            ll rev_target = graph[vertex][i].rev;
            // if capasitiy is not full yet and target is farther,
            // then assign current flow 
            if (cap_target > 0 && level[vertex] < level[target]) {
                ll d = dfs(target, sink, min(cap_target, flow));
                if (d > 0){ // if the flow successfully reaches the sink, reduce the flow from the capacity 
                    graph[vertex][i].capacity -= d;
                    graph[target][rev_target].capacity += d;
                    return d;
                }
            }
        }
        return 0;
    }

    ll dinic(ll source, ll sink) {
        ll flow = 0;
        while (true) {
            bfs(source);
            // if there is no path leading to the sink, the maximum flow is 0.
            if (level[sink] < 0) return flow;
            iter = vl(n_vertex, 0); ll f;
            while ((f = dfs(source, sink, INF)) > 0) flow += f;
        }
    }
};

class UnionFind {
    vl parents, set_size;
public:
    UnionFind(){}
    UnionFind(ll n){
        parents = set_size = vl(n);
        rep(i, n) {parents[i] = i; set_size[i] = 1LL;}
    }

    ll root_find(ll x){
        if (parents[x] == x) return x;
        return parents[x] = root_find(parents[x]);
    }

    void unite(ll x, ll y){
        // priority for x is larger than that of y
        x = root_find(x); y = root_find(y);
        if (x == y) return;
        if (set_size[x] < set_size[y]){
            parents[y] = x; set_size[x] += set_size[y];
        }else{
            parents[x] = y; set_size[y] += set_size[x];
        }
    }

    bool is_same(ll x, ll y) {// connected or not
        return root_find(x) == root_find(y);
    }

    ll size(ll x) {
        return set_size[root_find(x)];
    }
};

struct Doubling { // ABC167D
    ll n;
    ll sz;
    vector<vl> next; 
    /*
        next[k + 1][i] := next[k][next[k][i]]
        next[0][i] := edge[i]
        e.g. a0, a1, ..., an-1 / 0 <= ai <= n - 1
        a0 -> a[a0] -> a[a[a0]] -> ... -> a[a[...[a[0]]]] (N times)
        Let the function repeatedly input a[i] m times be f[m](a[i])
        
        - get(i, x) returns f[x](a[i])
        - lower_bound(i, j) returns minimum x which satisfies f[x](a[i]) >= j.
                                    if not possible returns n.

    */
 
    // edge[i]: the step size for one iteration
    Doubling(vl& edge) : n(edge.size()), sz(62) {
        next.resize(sz, vl(n, -1));
        rep(i, n) next[0][i] = edge[i];
        rep(k, sz - 1) rep(i, n) next[k + 1][i] = next[k][next[k][i]];
    }
    
    ll get(ll i, ll x) {
        ll ret = i;
        rep(bit, sz){
            if (!(x >> bit & 1)) continue;
            ret = next[bit][ret];
        }
        return ret;
    }
    
    ll lower_bound(ll i, ll j) {
        ll cur = i, acc = 0;
        revrep(wid, sz){
            if (next[wid][cur] < j) {
                acc += 1LL << wid;
                cur = next[wid][cur];
            }
        }
        return min(n, acc + 1);
    }
};

/*
class LCA{
public:
    ll N, logN;
    vl depth, len;
    Graph tree[200005]; // global declaration later.
    vector<vl> parents;

    LCA(ll n){
        N = n;
        logN = 0;
        while (N > (1LL << logN)) logN++;
        depth = vl(N); len = vl(N);
        parents = vector<vl>(logN, vl(N));
        init(0, -1, 0, 0);
        build();
    }

    void init(ll source, ll parent, ll d, ll l){
        depth[source] = d;
        parents[0][source] = parent;
        len[source] = l;
        rep(i, tree[source].size()){
            ll target = tree[source][i].to;
            ll cost = tree[source][i].cost;
            if (target == parent) continue;
            init(target, source, d + 1, cost + l);
        }
    }

    void build(){
        rep(k, logN - 1) rep(n, N){
            // if there is no parent, -1.
            // otherwise, the parent of the parent is the parent.
            if (parents[k][n] < 0) parents[k + 1][n] = -1;
            else parents[k + 1][n] = parents[k][parents[k][n]];
        }
    }

    ll query(ll u, ll v){
        if (depth[u] > depth[v]) swap(u, v);
        rep(k, logN) if ((depth[v] - depth[u]) >> k & 1) v = parents[k][v];
        if (u == v) return u;

        revrep(k, logN){
            if (parents[k][u] != parents[k][v]){
                u = parents[k][u]; v = parents[k][v];
            }
        }
        return parents[0][u];
    }

    ll distance(ll u, ll v){
        ll w = query(u, v);
        return len[u] + len[v] - 2 * len[w];
    }

};
*/

struct BIT{
    ll n; vl tree_dat;
    BIT(ll n): tree_dat(n + 1, 0), n(n){};

    //       x: 1001 1010  1100 1011 1101  1111
    // x & - x: 0001 0010  0100 0001 0001  0001
    //      ->: 1010 1100 10000 1100 1100 10000

    ll update_func(ll val, ll dat){
        // if maximum -> max(val, dat)
        // return max(val, dat);
        // if cumulative sum
        return val + dat;
    }

    ll query(ll i){
        /*
        e.g.) i = 10101
        itr1. 10101 -> 10100 
        itr2. 10100 -> 10000
        itr3. 10000 -> 00000 (break)
        */
        ll ret = 0;
        for (ll j = i; j >= 0; j = (j & (j + 1)) - 1){
            ret = update_func(ret, tree_dat[j]);
        }
        return ret;
    }

    ll lower_bound(ll key){
        if (key <= 0) return 0;
        ll left = 0, right = 1;
        while(right <= n) right *= 2;

        for (ll i = right; i > 0; i /= 2){
            if (left + i <= n && tree_dat[left + i - 1] < key){
                key -= tree_dat[left + i - 1];
                left += i;
            }
        }
        return left;
    }

    void update(ll i, ll val){
        /*
        e.g.) i = 10101, n = 11111
        itr1. 10101 ->  10110 
        itr2. 10110 ->  11000
        itr3. 11000 -> 100000 (break)
        */
        if (i < 0) return;
        for(ll j = i; j < n; j |= j + 1){
            tree_dat[j] = update_func(val, tree_dat[j]);
        }
    }
};

template<class Target>
class RerootingTreeDP {
public:
    using T = typename Target::type;
    struct DP_edge {
        ll to, rev; // rev is the index to trace the source node.
        T value; // objective value
    };
private:
    ll n;
    void dfs_fwd(ll source, ll parent){
        ll par_idx = -1;
        vector<T> values;
        rep(i, tree[source].size()) {
            const DP_edge& e = tree[source][i];
            if (e.to == parent){par_idx = i; continue;}
            dfs_fwd(e.to, source);
            values.pb(e.value);
        }
        
        // If the parent != -1, update the value on edge from parent to source
        if (par_idx != -1){
            ll src_idx = tree[source][par_idx].rev;
            // update values on the edge from parent to source
            tree[parent][src_idx].value = Target::merge(values);
        }
    }

    void dfs_bwd(ll source, ll parent){
        vector<T> values;
        for (auto&& e: tree[source]) values.pb(e.value);
        values = Target::evaluate(values);

        rep(i, tree[source].size()) {
            const DP_edge& e = tree[source][i];
            if (e.to == parent) continue;
            // tree[e.to][e.rev]: e.to -> source
            tree[e.to][e.rev].value = values[i];
            dfs_bwd(e.to, source);
        }
    }
public:
    UnionFind uf;
    vector<vector<DP_edge>> tree;
    RerootingTreeDP(ll n) : n(n), uf(n), tree(n) {}

    void add_edge(ll u, ll v, T val){
        assert(!uf.is_same(u, v));
        tree[u].pb({v, ll(tree[v].size()), val});
        tree[v].pb({u, ll(tree[u].size()) - 1, val});
        uf.unite(u, v);
    }
    
    void dp() {
        vector<bool> visited(n, false);
        rep(i, n){
            if (visited[uf.root_find(i)]) continue;
            dfs_fwd(i, -1);
            visited[uf.root_find(i)] = true;
        }

        visited.assign(n, false);
        rep(i, n){
            if (visited[uf.root_find(i)]) continue;
            dfs_bwd(i, -1);
            visited[uf.root_find(i)] = true;
        }
    }

    ll size() const {return tree.size();}
};

// ABC160F is one example
// Modify the functions evaluate and merge based on given problems
struct Merger {
    using type = ll;
  
    static vector<type> evaluate(const vector<type>& value){
        const ll n = value.size();
        vl L(n + 1, 1), R(n + 1, 1);
        rep(i, n)
            L[i + 1] = (L[i] * value[i]) % MOD;
        revrep(i, n)
            R[i] = (R[i + 1] * value[i]) % MOD;

        vl ret(n);
        /*
        ret[i] = v[c0] * v[c1] * ... * v[ci-1] * v[ci+1] * ... * v[cn-1]
        ci: the i-th child

        Let's consider the case where the root of a given tree is 0. (0-indexed)
        Here, we try to obtain the objective values for the edges from children to parent. 
        When we change the root of the tree into another, 
        the parent node here also changes the family structure.
        Since each node has ONLY one parent and zero or one or multiple children,
        the value for edge on child i to source is obtained using the value from source to child j.
        Here j is all the integers which is not i.

        In this case, child i is grandparent of child j for all j and
        the computation for the value from child i to source is O(N) when we use cumulative sum or product.
        */
        rep(i, n) ret[i] = (L[i] * R[i + 1]) % MOD;

        return ret;
    }

    static type merge(const vector<type>& value){
        ll ret;
        for (auto&& v: value){
        }
        return ret;
    }
};

ll gcd(ll m, ll n){
    ll a = max(m, n);
    ll b = min(m, n);
    while(b != 1 && b != 0){ a %= b; swap(a, b);}
    return b == 1 ? 1 : a;
}

ll lcm(ll m, ll n){ return m / gcd(m, n) * n;}

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {a = b; return true;}
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b){a = b; return true;}
    return false;
}

ll power_mod(ll a, ll power, ll mod){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = (value * a) % mod;
        a = (a * a) % mod;
        power = power >> 1;
    }
    return value % mod;
}

ll modinv(ll a, ll mod){
    return power_mod(a, mod - 2, mod);
}

ll power_normal(ll a, ll power){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = value * a;
        a = a * a;
        power = power >> 1;
    }
    return value;
}

ll comb_memo[55][55];

ll pascal_triangle(ll n){
    comb_memo[0][0] = 1;
    For(i, 1, n + 1) rep(j, i + 1){
        comb_memo[i][j] += comb_memo[i - 1][j];
        if (j > 0) comb_memo[i][j] += comb_memo[i - 1][j - 1];
    }
}

ll combination(ll n, ll r, ll mod){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = 1;
    ll denomenator = 1;
    for(ll i = 0; i < r ; i++){
        ll num = (n - i) % mod, den = (i + 1) % mod;
        (numerator *= num) %= mod;
        (denomenator *= modinv(den, mod)) %= mod;
    }

    return (numerator * denomenator) % mod;
}

ll combination_no_mod(ll n, ll r){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = 1;
    ll denomenator = 1;
    for(ll i = 0; i < r ; i++){
        numerator *= n - i;
        denomenator *= i + 1;
        ll g = gcd(numerator, denomenator);
        numerator /= g; denomenator /= g;
    }

    return numerator;
}

ld log_combination(ll n, ll r){
    if (n == r && n == 0) return 0;
    else if (n <= 0 || r < 0 || r > n) return -INF;
    ld val = 0;
    for(ll i = 0; i < r ; i++){
        val += log(n - i);
        val -= log(i + 1);
    }

    return val;
}

ll bin_search(ll key, ll A[], ll left, ll right){
    // return the index idx where A[idx] = key.
    // A[left] is start and A[right] is end..
    // In other words, A[right], not A[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (A[mid] == key) return mid;
        else if (A[mid] > key) right = mid - 1;
        else if (A[mid] < key) left = mid + 1;
    }
    return -1;
}

/*
ll bin_search_temp(ll left, ll right, callable judge){
    while(right > left){
        // when seeking lower bound
        ll mid = (right + left) / 2;
        if (judge(mid)) right = mid;
        else left = mid + 1;
        // when seeking upper bound
        ll mid = (right + left + 1) / 2;
        if (judge(mid)) left = mid;
        else right = mid - 1;
    }

    return right;
}

trinary_search

// Care the value EPS!!! Compare to the condition
ld left = 0; ld right = p;
while(abs(right - left) > EPS){
    ld left2 = (2 * left + right) / 3.0;
    ld right2 = (left + 2 * right) / 3.0;
    ld f1 = tak_func(left2);
    ld f2 = tak_func(right2);
    if (f1 <= f2) right = right2;
    else if (f2 <= f1) left = left2;
}

*/

str bin_expression(ll n, ll dig){
    str s = "";
    rep(i, dig){
        s += ltos(n % 2);
        n /= 2;
    }
    reverse(all(s));
    return s;
}

ll lower_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] >= key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] < key && key <= A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] < key && key <= A[mid]) return mid;
        else if (A[mid - 1] >= key) right = mid - 1;
        else if (A[mid] < key) left = mid;
    }

    return -1; // all the elements < key
}

ll inf_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] > key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] <= key && key < A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] <= key && key < A[mid]) return mid;
        else if (A[mid - 1] > key) right = mid - 1;
        else if (A[mid] <= key) left = mid;
    }

    return -1; // all the elements <= key
}

ll upper_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] <= key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] <= key && key < A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] <= key && key < A[mid + 1]) return mid;
        else if (A[mid] > key) right = mid;
        else if (A[mid + 1] <= key) left = mid + 1;
    }

    return -1; // all the elements > key
}

ll sup_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] < key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] < key && key <= A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] < key && key <= A[mid + 1]) return mid;
        else if (A[mid] >= key) right = mid;
        else if (A[mid + 1] < key) left = mid + 1;
    }

    return -1; // all the elements >= key
}

ll bin_search_vector(ll key, vl v, ll left, ll right){
    // return the index idx where v[idx] = key.
    // v[left] is start and v[right] is end..
    // In other words, v[right], not v[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (v[mid] == key) return mid;
        else if (v[mid] > key) right = mid - 1;
        else if (v[mid] < key) left = mid + 1;
    }
    return -1;
}

ll lower_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] < key <= v[N]
    // N == -1 if all the elements < key
    return lower_bound(all(v), key) - v.begin();
}

ll inf_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] <= key < v[N] <= key + 1
    // N == -1 if all the elements <= key
    return lower_bound(all(v), key + 1) - v.begin();
}

ll upper_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key < v[N + 1]
    // N == -1 if all the elements > key
    return upper_bound(all(v), key) - v.begin(); // (- 1)
}

ll sup_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key - 1 < key <= v[N + 1]
    // N == -1 if all the elements >= key
    return upper_bound(all(v), key - 1) - v.begin() - 1;
}

ll fact(ll n){
    if (n == 0) return 1;
    return n * fact(n - 1);
}

ll fact_mod(ll n, ll mod){
    if (n == 0) return 1;
    return n * fact_mod(n - 1, mod);
}

/*
ll facts[6000], facts_inv[6000], inv[6000];

void fact_memo(ll n, ll mod){
    facts[0] = facts[1] = 1;
    facts_inv[0] = facts_inv[1] = 1;
    inv[1] = 1;
    
    For(i, 2, n + 1){
        facts[i] = (i * facts[i - 1]) % mod;
        inv[i] = mod - inv[mod % i] * (mod / i) % mod;
        facts_inv[i] = facts_inv[i - 1] * inv[i] % mod;
    }
}

ll nCk(ll n, ll r){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll val = (facts[n] * facts_inv[n - r]) % MOD;
    val *= facts_inv[r];
    return val % MOD;
}
*/

bool is_prime(ll n){
    if (n <= 1) return false;
    for(ll i = 2; i * i <= n; i++){
        if (n % i == 0) return false;
    }
    return true;
}

ll bool_sum(ll a1, ll a2){
    if (a1 == 1 || a2 == 1) return 1;
    return 0;
}

mll prime_factorization(ll n){
    ll i = 2;
    mll table;
    while(i * i <= n){
        while(n % i == 0){
            table[i] ++;
            n /= i;
        }
        i++;
    }
    if (n > 1) table[n] = 1;
    return table;
}

vl divisor_table(ll n){
    vl table;
    ll i = 1;
    while(i * i <= n){
        if (n % i == 0){
            table.pb(i);
            if (i * i != n) table.pb(n / i);
        }
        i++;
    }

    sort(all(table));

    return table;
}

ll char_to_idx(char c){
    ll idx = 0;
    Forchar(cc, 'a', 'z'){
        if (c == cc) return idx;
        else idx ++;
    }
}

ll next_combination(ll sub){
    /*
    ### Attention ###
    if k is 0 or 1 and n is 0, it does not work well.

    ll n, k; ll bit = (1 << k) - 1;
    for (; bit < (1 << n); bit = next_combination(bit)){
        bool ith = bit & (1 << i);
        procedures...
    }

        sub & -sub: the binary which shares the last digit whose value is 1 in sub
        sub + x   : carry up the last digit
            ~y    : the binary whose digits are 1 if y's digit is 0.
    (sub & ~y) / x: reduce the same number of 0s after first 1 in x from (sub & ~y).
    */
    ll x = sub & -sub, y = sub + x;
    return (((sub & ~y) / x) >> 1) | y; 
}

vl z_algorithm(str s){
    ll n = s.length(); vl res(n);
    res[0] = n;
    ll i1 = 1, i2 = 0;

    while(i1 < n){
        while (i1 + i2 < n && s[i2] == s[i1 + i2]) ++i2;
        res[i1] = i2; if (i2 == 0){++i1; continue;}
        ll i3 = 1;
        while (i1 + i3 < n && i3 + res[i3] < i2){res[i1 + i3] = res[i3]; ++i3;}
        i1 += i3, i2 -= i3;
    }
    return res;
}

ll big_number_mod(str s, ll mod){
    ll l = s.length(); ll idx = 0;
    ll val = 0; ll tenth = 1;
    while (idx < l){
        ll m = ctol(s[l - 1 - idx]);
        val += (m * tenth) % mod; val %= mod;
        tenth *= 10; tenth %= mod; idx++;
    }

    return val;
}

ll string_to_ll(str s){
    ll l = s.length(); ll idx = 0;
    ll val = 0; ll tenth = 1;
    while (idx < l){
        ll m = ctol(s[l - 1 - idx]);
        val += (m * tenth);
        tenth *= 10; idx++;
    }

    return val;
}

str reflected_string(str s){
    str t, u; ll n = s.length();
    t = s; reverse(all(t));
    u = substring(t, 0, n - 2) + s + substring(t, 1, n - 1);
    return u;
}

ld distance_between_point_line(point l_begin, point l_end, point p){
    ll xl1 = l_begin.x, yl1 = l_begin.y; 
    ll xl2 = l_end.x, yl2 = l_end.y;
    ll xp = p.x, yp = p.y;
    ll a = yl2 - yl1;
    ll b = - xl2 + xl1;
    ll c = - a * xl2 - b * yl2;
    return abs(ld(a * xp + b * yp + c)) / ld(sqrt(a * a + b * b));
}

bool is_cross(point l1_begin, point l1_end, point l2_begin, point l2_end){
    ll x1 = l1_begin.x, y1 = l1_begin.y;
    ll x2 = l1_end.x, y2 = l1_end.y;
    ll x3 = l2_begin.x, y3 = l2_begin.y;
    ll x4 = l2_end.x, y4 = l2_end.y;
    ll val1 = (x1 - x2) * (y3 - y1) + (y1 - y2) * (x1 - x3);
    ll val2 = (x1 - x2) * (y4 - y1) + (y1 - y2) * (x1 - x4);
    ll val3 = (x3 - x4) * (y1 - y3) + (y3 - y4) * (x3 - x1);
    ll val4 = (x3 - x4) * (y2 - y3) + (y3 - y4) * (x3 - x2);

    return val1 * val2 < 0 && val3 * val4 < 0;
}

ld space_of_triangle(point p1, point p2, point p3){
    ll x1 = p1.x, y1 = p1.y;
    ll x2 = p2.x, y2 = p2.y;
    ll x3 = p3.x, y3 = p3.y;
    ll v1 = x2 - x1; ll u1 = y2 - y1;
    ll v2 = x3 - x1; ll u2 = y3 - y1;
    ld s = ld(v1 * u2 - u1 * v2) / ld(2);
    return abs(s);
}

pair<point, ll> center_2det_of_3points(point p1, point p2, point p3){
    // the center of circle on the given three points
    // return the determinant value and the product of center points and 2 * determinant value
    ll x1 = p1.x, y1 = p1.y;
    ll x2 = p2.x, y2 = p2.y;
    ll x3 = p3.x, y3 = p3.y;

    // center of 2 points
    ll c2x_2 = x2 + x1, c2y_2 = y2 + y1;
    ll c3x_2 = x3 + x1, c3y_2 = y3 + y1;

    // vertical vector of 2 lines
    ll b2y = x2 - x1, b2x = - y2 + y1;
    ll b3y = x3 - x1, b3x = - y3 + y1;

    ll x1_2 = c3x_2 * b3y - c3y_2 * b3x, y1_2 = c2x_2 * b2y - c2y_2 * b2x;
    ll det = - b3y * b2x + b3x * b2y;
    if (det == 0) return {{INF, INF}, det};
    ll cx_2det = - b2x * x1_2 + b3x * y1_2, cy_2det = - b2y * x1_2 + b3y * y1_2;
    return {{cx_2det, cy_2det}, det};
}

ll inversion_number(vl a, ll a_max){
    /*
    Paramters
    ---------
    a: vector<ll>
        All the elements must be non-negative.
        Prefably the elements are compressed to reduce the computational cost.
    a_max: ll
        The maximum value of the vector a or the value bigger than the value stated previously.
    */

    BIT bit(a_max + 1); ll val = 0;
    rep(i, a.size()){
        // i is the number of elements that have lower index than a[i].
        // call the number of elements that have lower value than a[i]
        // by subtracting these two, the residual number is the number of elements that have larger value 
        val += i - bit.query(a[i] - 1);// cumulative sum from 0 to a[i] - 1
        bit.update(a[i], 1);
    }
    return val;
}

template<typename T>
vector<T> compress(vector<T> v){
    // sort and remove all the duplicated values
    sort(all(v));
    v.erase(unique(all(v)), v.end());
    return v;
}

template<typename T>
map<T, ll> dict(const vector<T> &v){
    map<T, ll> d;
    rep(i, v.size()) d[v[i]] = i;
    return d;
}

void GaussJordanBitVector(vl &bs){
    ll n = bs.size();
    ll rank = 0; ll j = 0;
    revrep(i, N_DIGITS){
        for(j = rank; j < n; j++) if (bs[j] & (1LL << i)) break;
        if (j == n) continue;
        if (j > rank) bs[rank] ^= bs[j];
        for (j = rank + 1; j < n; j++) bs[j] = min(bs[j], bs[j] ^ bs[rank]);
        rank++;
    }
}

ll kruskal(vector<undirected_edge>& es, ll n_vertex){
    sort(all(es));
    UnionFind uf(n_vertex);

    ll min_cost = 0;

    rep(i, es.size()){
        undirected_edge& e = es[i];
        if (!uf.is_same(e.from, e.to)){
            min_cost += e.cost;
            uf.unite(e.from, e.to);
        }
    }
    return min_cost;
}

/*
const ll N_VERTEX = 310;
ll a, b, t;
ll dist[N_VERTEX][N_VERTEX];

void warshall_floyd(ll n){
    // rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(k, n) rep(i, n) rep(j, n) dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
}

int main(void){
    in2(n, m);
    rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        dist[a][b] = t;
        dist[b][a] = t;
    }

    warshall_floyd(n);
}

ll dist[1005], vertex_pre[1005];

void dijkstra(ll start, ll n) {
    priority_queue<pl, vector<pl>, greater<pl>> edge_costs;
    fill(dist, dist + n, INF); // fill(vertex_pre, vertex_pre + n, -1);
    dist[start] = 0;
    edge_costs.push(pl(0, start));

    while (!edge_costs.empty()) {
        pl edge_cost = edge_costs.top();
        edge_costs.pop();
        ll idx = edge_cost.second;
        ll cost = edge_cost.first;
        if (dist[idx] < cost) continue;

        rep(i, graph[idx].size()){
            edge e = graph[idx][i];
            if (dist[e.to] > dist[idx] + e.cost){
                dist[e.to] = dist[idx] + e.cost;
                // vertex_pre[e.to] = idx;
                edge_costs.push(pl(dist[e.to], e.to));
            }
        }
    }
}

vl get_predecessor(ll g){
    vl path;
    for (; g != -1; g = vertex_pre[g]) path.pb(g);
    
    reverse(all(path));
    return path;
}

int main(void){
    in2(n, m);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        G[a].pb({b, t});
        G[b].pb({a, t});
    }

    dijkstra(0, n);
}


# ABC061D

bool find_negative_cycle(ll goal){
    rep(i, n) rep(v, n) rep(k, graph[v].size()){
        edge e = graph[v][k];
        if (dist[e.to] != INF && dist[e.to] > dist[v] + e.cost){
            dist[e.to] = -INF;
            if (goal == -1) return true;
            else if (goal == e.to) return true;
        }
    }
    return false;
}

bool bellman_ford(ll start, ll n, ll goal){
    // if there is a closed circuit, it returns false. (when goal == -1)
    // if the distance to goal cannot be obtained, it returns false (when goal != -1)
    fill(dist, dist + n, INF);
    dist[start] = 0;
    rep(i, n) rep(v, n) rep(k, graph[v].size()){
        edge e = graph[v][k];
        if (dist[v] != INF && dist[e.to] > dist[v] + e.cost) 
            dist[e.to] = dist[v] + e.cost;
    }

    if (find_negative_cycle(goal)) return false;
    return true;
}

*/

/*
    # 1. The usage of map pair
    map<pl, ll> cnt;
    cnt[{i, j}] = 0;
    items(kv, cnt){
        prl2(kv.first, kv.second);
    }

    # 2. The usage of next_permutation and combination (factorial search)
    ll a[8];
    rep(i, 8) a[i] = i;
    sort(a, a + 8);
    do{
    }while(next_permutation(a, a+n));

    // here, combination
    ll n, r;
    ll i1, i2, ..., ir;
    For(i1, n - r, n) For(i2, n - r - 1, i1) ... For(ir, n - 2 * r + 1, irr){ process;}

    # 3. bit search
    ll n;
    in1(n);
    const ll base = 3;
    ll upper = power_normal(base, n);
    
    rep(i, upper){
        ll tmp = i;
        rep(j, n){
            rep(k, base) if (tmp % base == k) prl(k);
            tmp /= base;
        }
    }

    # 4. imos method
    // used when we would like to count the number which
    // shows how many times the numbers between l and r belongs to smt.
    // This method is composed of three process.
    ll n, m, s[MAX_M], l, r;
    in2(n, m);
    rep(i, m) s[i] = 0;

    // 1st step
    rep(i, n){
        in3(l, r, c);
        l--; r--; // if l starts from 1.
        s[l] += c; s[r + 1] -= c;
    }

    // 2nd step
    rep(i, m - 1) s[i + 1] += s[i];
    // 3rd step: judgement...

    #5. shakutori method (syakutori, two pointers technique)
    // 1. strech right side while the condition is met.
    // 2. renew the answer
    // 3. increments left side
    // 4. Back to 1. (l <= r must be satisfied all the time.)

    ll l = 0; ll r = 0;
    while (l < n){
        r = max(r, l);
        if (l == r) r++;
        while(r < n && cond) r++;
        answer += r - l; l++;
    }
    prl(answer);

    #6. priority queue

    pq q;
    ll answer = 0;
    ll v;

    rep(i, n) q.push(a[i]);
    rep(i, m){
        v = q.top(); q.pop(); // get the top value and dump the value from queue
        v /= 2; q.push(v); // add the new value
    }

    while(!q.empty()){
        answer += q.top();
        q.pop();
    }

    #7. The shortest path (distance) between the k-th edge and another edge (Tree)
    Graph tree[MAX_N];
    ll depth[MAX_N];

    void dfs(ll source, ll parent, ll all_cost){
        depth[source] = all_cost;
        items(e, tree[source]){
            if (e.to == parent) continue;
            dfs(e.to, source, all_cost + e.cost);
        }
    }

    ll n, k, a, b, c;

    in2(n, k);
    rep(i, n - 1){
        in3(a, b, c);
        a--; b--;
        tree[a].pb({b, c});
        tree[b].pb({a, c});
    }

    k--;
    dfs(k, -1, 0);

    #10. Visiting Subtree using recurrent function (ABC138D)

    Graph tree[MAX_N];
    ll c[MAX_N];
    bool visited[MAX_N];

    void dfs(ll source, ll parent, ll val){
        visited[source] = true;
        c[source] += val;
        rep(i, tree[source].size()){
            rep(i, m){ll res = n % match[i].e1;}
            ll vertex = tree[source][i].to;
            if (vertex == parent) continue;
            dfs(vertex, source, c[source]);
        }
    }

    #11. bfs ABC146D, ABC007C
    1. first create a tree.
    2. start searching from a node.
    3. do some processes and push nodes connected with a given target node in BFS.
    4. repeat a series of procedure until queue is empty.

    queue<pl> q;

    void bfs(ll source, ll parents){
        ll n_edge = graph[source].size();
        if (parents != -1) dist[source] = min(dist[source], dist[parents] + 1);
        if (visited[source]) return;
        visited[source] = true;

        rep(idx, n_edge){
            ll target = graph[source][idx].to;
            if (target == parents) continue;
            q.push(mp(target, source)); 
        }
    }

    q.push(mp(sg.e1, -1));
    while(!q.empty()){
        pl source = q.front(); q.pop();
        bfs(source.e1, source.e2);
    }

    #12. grid to distance matrix (dx, dy)
    ll w, h;
    ll pos_to_idx(ll x, ll y){
        return y * w + x;
    }

    pl idx_to_pos(ll idx){
        return mp(idx % w, idx / w); 
    }

    int main(void){
        in2(h, w);
        rep(y, h){
            in1(s);
            rep(x, w){
                if (s[x] == '#') wall[x][y] = true;
                else wall[x][y] = false;
            }
        }

        rep(i1, h * w)rep(i2, h * w) dist[i1][i2] = INF * (i1 != i2);

        rep(x, w)rep(y, h){
            ll idx1 = pos_to_idx(x, y); ll idx2;
            if (wall[x][y]) continue;
            if (x != 0 && !wall[x - 1][y]){
                idx2 = pos_to_idx(x - 1, y);
                // if warshall floyd
                dist[idx1][idx2] = 1;
                // if dijkstra
                // graph[idx1].pb({idx2, 1});
            }
            if (x != w - 1 && !wall[x + 1][y]){
                idx2 = pos_to_idx(x + 1, y);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != 0 && !wall[x][y - 1]){
                idx2 = pos_to_idx(x, y - 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != h - 1 && !wall[x][y + 1]){
                idx2 = pos_to_idx(x, y + 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
        }
    }

    # Cumulative sum (2 dimension)

    ll func(ll x, ll y, ll dx, ll dy){
        if (x + dx > w || y + dy > h) return - INF;
        ll val = cum[x + dx][y + dy];
        val += cum[x][y];
        val -= cum[x][y + dy];
        val -= cum[x + dx][y];
        return val;
    }

    rep(x, w + 1) cum[x][0] = 0;
    rep(y, h + 1) cum[0][y] = 0;
    rep(y, h + 1) rep(x, w)
        cum[x + 1][y + 1] = cum[x][y + 1] + vs[x][y];

    rep(x, w + 1) rep(y, h)
        cum[x][y + 1] += cum[x][y];
    
    # DEBUG_MATRIX

    rep(i, n){
        rep(j, n) prs(a[i][j]);
        prl("");
    }

*/

/*
# the operators regarding bit
& (AND), | (OR), ^ (XOR)
- (REVERSE), >> (SMALLER SHIFT)
<< (BIGGER SHIFT)


     x1: 0000 0001 0010 0101 0110 0111 0111
     x2: xxxx 0001 0011 0100 0101 1000 0110
x1 & x2: 0000 0001 0010 0100 0100 0000 0110

      x: 1001 1010  1100 1011 1101  1111
x & - x: 0001 0010  0100 0001 0001  0001
    sum: 1010 1100 10000 1100 1100 10000

x << y is x * 2 ** y
x >> y is rep(i, y) x = x // 2 

####### Attention #######
1 << i and 1LL << i is different. If programs show WA, try switch to this one.
Let S be a bit sequence and i be a non-negative integer
S & (1 << i) -> if true, i in S
S | (1 << i) -> S union {i}
S & -(1 << i) -> S - {i}
__builtin_popcount(S) -> the number of elements in S
S = 0 -> S is an empty set
__builtin_popcountl(i) -> the number of 1 in binary
S = (1 << n) - 1 -> S includes all the elements up to the n-th

#Conditional Operator
condition ? true : false;

#iterator
type declaration: auto
value reference: *itr
increment: itr++
decrement: itr--
substitution of value: *itr = smt

# inclusion-exclusion principle
|U[i = 1 to n] Ai| = sum[i = 1 to n] |Ai| - sum[i < j]|Ai ^ Aj| + ... + (-1)^(n - 1) |^[i = 1 to n]Ai|
*/

const ll MAX_N = 200005;
bool okay = false;
ll answer = 0;
ll n, k, nz;
vl as, pos, neg;

ll func(ll x){
    ll cnt = 0;
    if (x >= 0) cnt += nz * (n - 1);
    
    ll pre = cnt;
    rep(i, pos.size()){
        ll target = floor(ld(x) / ld(pos[i]));
        cnt += upper_bound(all(as), target) - as.begin();
        if (pos[i] * pos[i] <= x) cnt--;
    }
    
    rep(i, neg.size()){
        ll target = floor(ld(- x - 1) / ld(- neg[i]));
        cnt += n - (upper_bound(all(as), target) - as.begin());
        if (neg[i] * neg[i] <= x) cnt--;
    }
 
    return cnt / 2;
}

void solve(){
    as = vl(n); nz = 0;
    rep(i, n) in1(as[i]);
    sort(all(as)); 
    
    rep(i, n){
        if (as[i] == 0) nz++;
        else if (as[i] > 0) pos.pb(as[i]);
        else if (as[i] < 0) neg.pb(as[i]);
    }

    ll left = -INF, right = INF;
    while (right > left + 1){
        ll mid = (left + right) / 2;
        if (func(mid) >= k) right = mid;
        else left = mid;
    }

    prl(right);
}

int main(void){
    in2(n, k);
    solve();
    // assert(n <= 10000);
    return 0;
}

Measured memory_bytes: 4084