Problem: You are given a binary array A=(A_1,A_2,\cdots,A_N) of length N.

Process Q queries of the following types. The i-th query is represented by three integers T_i,L_i,R_i.

* T_i=1: Replace the value of A_j with 1-A_j for each L_i \leq j \leq R_i.
* T_i=2: Calculate the inversion(*) of the array A_{L_i},A_{L_i+1},\cdots,A_{R_i}．



Note：The inversion of the array x_1,x_2,\cdots,x_k is the number of the pair of integers i,j with 1 \leq i < j \leq k, x_i > x_j.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 0 \leq A_i \leq 1
* 1 \leq Q \leq 2 \times 10^5
* 1 \leq T_i \leq 2
* 1 \leq L_i \leq R_i \leq N
* All values in Input are integer.

Input

Input is given from Standard Input in the following format:


N Q
A_1 A_2 \cdots A_N
T_1 L_1 R_1
T_2 L_2 R_2
\vdots
T_Q L_Q R_Q


Output

For each query with T_i=2, print the answer.

Example

Input

5 5
0 1 0 0 1
2 1 5
1 3 4
2 2 5
1 1 3
2 1 2


Output

2
0
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) { cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl; }
#define mdebug(m) { cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;} }
#define pb push_back
#define fi first
#define se second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p){ cout << '(' << p.first << ',' << p.second << ')'; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void chOut(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using v_bool = vector<bool>;
using v_Pii = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;

struct LazySegmentTree
{
    //A: data, B: 作用素
    using T_A = PiP;
    using T_B = bool;
    T_A id_A = PiP(0, Pii(0, 0));
    T_B id_B = false;
    int N;
    vector<T_A> dat;
    vector<T_B> ope;

    T_A F_AA(T_A &u, T_A &v){
        int n0 = u.fi, a0 = u.se.fi, b0 = u.se.se;
        int n1 = v.fi, a1 = v.se.fi, b1 = v.se.se;
        return PiP(n0 + n1 + b0 * a1, Pii(a0 + a1, b0 + b1));
    }

    T_A F_AB(T_A &u, T_B v, int sz){
        if(v){
            int n = u.fi, a = u.se.fi, b = u.se.se;
            return PiP(a * b - n, Pii(b, a));
        }else return u;
    }

    T_B F_BB(T_B u, T_B v){
        return u ^ v;
    }

    LazySegmentTree(int n){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
    }

    LazySegmentTree(int n, vector<T_A> &v){
        N = 1;
        while(n > N) N = N << 1;
        dat = vector<T_A>(2 * N - 1, id_A);
        ope = vector<T_B>(2 * N - 1, id_B);
        REP(i, n) dat[i + N - 1] = v[i];
        IREP(i, N - 1) dat[i] = F_AA(dat[i * 2 + 1], dat[i * 2 + 2]); 
    }

    //遅延評価
    void eval(int k, int l, int r){
        if(ope[k] != id_B){
            dat[k] = F_AB(dat[k], ope[k], r - l);
            if(r - l > 1){
                ope[k * 2 + 1] = F_BB(ope[k * 2 + 1], ope[k]);
                ope[k * 2 + 2] = F_BB(ope[k * 2 + 2], ope[k]);
            }
            ope[k] = id_B;
        }
    }

    void update(int a, int b, T_B x, int k, int l, int r){

        eval(k, l, r);
        if(b <= l || r <= a) return;

        if(a <= l && r <= b){
            ope[k] = F_BB(ope[k], x);
            eval(k, l, r);
        }else{
            update(a, b, x, k * 2 + 1, l, (l + r) / 2);
            update(a, b, x, k * 2 + 2, (l + r) / 2, r);
            dat[k] = F_AA(dat[k * 2 + 1], dat[k * 2 + 2]);
        }

    }

    T_A get(int a, int b, int k, int l, int r){
        if(r <= a || b <= l) return id_A;

        eval(k, l, r);
        if(a <= l && r <= b) return dat[k];
        else{
            T_A vl = get(a, b, k * 2 + 1, l, (l + r) / 2);
            T_A vr = get(a, b, k * 2 + 2, (l + r) / 2, r);
            return F_AA(vl, vr);
        }
    }

    //[a,b)の区間更新
    void update(int a, int b, T_B x){ update(a, b, x, 0, 0, N); }

    //[a, b)の演算結果
    T_A get(int a, int b){ return get(a, b, 0, 0, N); }

    T_A val(int k){ return get(k, k + 1); }
};

signed main(){

    int N, Q; cin >> N >> Q;
    vec A(N); cin >> A;
    
    vector<PiP> p(N);
    REP(i, N){
        if(A[i] == 0) p[i] = PiP(0, Pii(1, 0));
        else p[i] = PiP(0, Pii(0, 1));
    }
    LazySegmentTree ST(N, p);

    REP(_, Q){
        int t, l, r; cin >> t >> l >> r;
        l--; r--;
        if(t == 1){
            ST.update(l, r + 1, true);
        }else{
            PiP res = ST.get(l, r + 1);
            Out(res.fi);
        }
    }

    return 0;
}