Problem: Polycarp plays a computer game. In this game, the players summon armies of magical minions, which then fight each other.

Polycarp can summon n different minions. The initial power level of the i-th minion is a_i, and when it is summoned, all previously summoned minions' power levels are increased by b_i. The minions can be summoned in any order.

Unfortunately, Polycarp cannot have more than k minions under his control. To get rid of unwanted minions after summoning them, he may destroy them. Each minion can be summoned (and destroyed) only once.

Polycarp's goal is to summon the strongest possible army. Formally, he wants to maximize the sum of power levels of all minions under his control (those which are summoned and not destroyed).

Help Polycarp to make up a plan of actions to summon the strongest possible army!

Input

The first line contains one integer T (1 ≤ T ≤ 75) — the number of test cases.

Each test case begins with a line containing two integers n and k (1 ≤ k ≤ n ≤ 75) — the number of minions availible for summoning, and the maximum number of minions that can be controlled by Polycarp, respectively.

Then n lines follow, the i-th line contains 2 integers a_i and b_i (1 ≤ a_i ≤ 10^5, 0 ≤ b_i ≤ 10^5) — the parameters of the i-th minion.

Output

For each test case print the optimal sequence of actions as follows:

Firstly, print m — the number of actions which Polycarp has to perform (0 ≤ m ≤ 2n). Then print m integers o_1, o_2, ..., o_m, where o_i denotes the i-th action as follows: if the i-th action is to summon the minion x, then o_i = x, and if the i-th action is to destroy the minion x, then o_i = -x. Each minion can be summoned at most once and cannot be destroyed before being summoned (and, obviously, cannot be destroyed more than once). The number of minions in Polycarp's army should be not greater than k after every action.

If there are multiple optimal sequences, print any of them.

Example

Input


3
5 2
5 3
7 0
5 0
4 0
10 0
2 1
10 100
50 10
5 5
1 5
2 4
3 3
4 2
5 1


Output


4
2 1 -1 5
1
2
5
5 4 3 2 1

Note

Consider the example test.

In the first test case, Polycarp can summon the minion 2 with power level 7, then summon the minion 1, which will increase the power level of the previous minion by 3, then destroy the minion 1, and finally, summon the minion 5. After this, Polycarp will have two minions with power levels of 10.

In the second test case, Polycarp can control only one minion, so he should choose the strongest of them and summon it.

In the third test case, Polycarp is able to summon and control all five minions.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
constexpr long double m_pi = 3.1415926535897932L;
constexpr long long MOD = 1000000007;
constexpr long long INF = 1LL << 61;
constexpr long double EPS = 1e-10;
template <typename T>
using vector2 = vector<vector<T>>;
template <typename T>
using vector3 = vector<vector2<T>>;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  int to, rev;
  long long cap;
  Edge(int _to, long long _cap, int _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, int from, int to, long long cap,
              bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size()));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (long long(i) = (long long)(0); i < (long long)(used.size()); i++)
      used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    negative[i] = false;
  d[s] = 0;
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (long long(i) = (long long)(0); i < (long long)(G[a.second].size());
         i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i].resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(d[i].size()); j++) {
      d[i][j] = ((i != j) ? INF : 0);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
      chmin(d[i][G[i][j].to], G[i][j].cap);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G.size()); j++) {
      for (long long(k) = (long long)(0); k < (long long)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<long long>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (long long(i) = (long long)(0); i < (long long)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class Lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  Lca() {}
  Lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
void visit(const vector<vector<Edge>>& g, int v, vector<vector<int>>& scc,
           stack<int>& S, vector<int>& inS, vector<int>& low, vector<int>& num,
           int& time) {
  low[v] = num[v] = ++time;
  S.push(v);
  inS[v] = true;
  for (decltype((g[v]).begin()) e = (g[v]).begin(); e != (g[v]).end(); ++e) {
    int w = e->to;
    if (num[w] == 0) {
      visit(g, w, scc, S, inS, low, num, time);
      low[v] = min(low[v], low[w]);
    } else if (inS[w])
      low[v] = min(low[v], num[w]);
  }
  if (low[v] == num[v]) {
    scc.push_back(vector<int>());
    while (1) {
      int w = S.top();
      S.pop();
      inS[w] = false;
      scc.back().push_back(w);
      if (v == w) break;
    }
  }
}
void stronglyConnectedComponents(const vector<vector<Edge>>& g,
                                 vector<vector<int>>& scc) {
  const int n = g.size();
  vector<int> num(n), low(n);
  stack<int> S;
  vector<int> inS(n);
  int time = 0;
  for (long long(u) = (long long)(0); u < (long long)(n); u++)
    if (num[u] == 0) visit(g, u, scc, S, inS, low, num, time);
}
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  long long size(int x) { return -data[root(x)]; }
  long long numSet() { return num; }
};
template <typename T, typename F>
class SegmentTree {
 private:
  T identity;
  F merge;
  long long n;
  vector<T> dat;

 public:
  SegmentTree(F f, T id, vector<T> v) : merge(f), identity(id) {
    int _n = v.size();
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
    for (long long(i) = (long long)(0); i < (long long)(_n); i++)
      dat[n + i - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
  }
  SegmentTree(F f, T id, int _n) : merge(f), identity(id) {
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
  }
  void set_val(int i, T x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  T query(int l, int r) {
    T left = identity, right = identity;
    l += n - 1;
    r += n - 1;
    while (l < r) {
      if ((l & 1) == 0) left = merge(left, dat[l]);
      if ((r & 1) == 0) right = merge(dat[r - 1], right);
      l = l / 2;
      r = (r - 1) / 2;
    }
    return merge(left, right);
  }
};
template <typename T>
class FenwickTree {
  vector<T> data;
  int n;
  int p;

 public:
  FenwickTree(int n) : n(n) {
    data.resize(n + 1LL, 0);
    p = 1;
    while (p < data.size()) p *= 2;
  }
  T sum(int k) {
    T ret = 0;
    for (; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }
  T sum(int a, int b) { return sum(b) - sum(a); }
  void add(int k, T x) {
    for (++k; k <= n; k += k & -k) data[k] += x;
  }
  int lower_bound(long long w) {
    if (w <= 0) return -1;
    int x = 0;
    for (int k = p / 2; k > 0; k /= 2) {
      if (x + k <= n && data[x + k] < w) w -= data[x + k], x += k;
    }
    return x;
  }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
void prime_factorization(long long n, vector<pair<long long, long long>>& ret) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back({i, 0});
      while (n % i == 0) {
        n /= i;
        ret[ret.size() - 1].second++;
      }
    }
  }
  if (n != 1) ret.push_back({n, 1});
}
long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> inv;
  long long mod;
  long long mod_inv(long long x) {
    long long n = mod - 2LL;
    long long res = 1LL;
    while (n > 0) {
      if (n & 1) res = res * x % mod;
      x = x * x % mod;
      n >>= 1;
    }
    return res;
  }
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    if (n < mod) return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
    long long ret = 1;
    while (n || r) {
      long long _n = n % mod, _r = r % mod;
      n /= mod;
      r /= mod;
      (ret *= nCr(_n, _r)) %= mod;
    }
    return ret;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long _n, long long _mod) {
    mod = _mod;
    long long n = min(_n + 1, mod);
    fact.resize(n);
    fact[0] = 1;
    for (long long(i) = (long long)(0); i < (long long)(n - 1); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    inv.resize(n);
    inv[n - 1] = mod_inv(fact[n - 1]);
    for (int i = n - 1; i > 0; i--) {
      inv[i - 1] = inv[i] * i % mod;
    }
  }
};
long long popcount(long long x) {
  x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
  x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
  x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
  x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
  x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
  x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);
  return x;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  using T = tuple<long long, long long, long long>;
  vector<T> m(n);
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    long long a, b;
    cin >> a >> b;
    m[i] = {b, a, i + 1};
  }
  sort((m).begin(), (m).end());
  vector3<long long> dp(
      n + 1, vector2<long long>(k + 1, vector<long long>(n - k + 1, -INF)));
  dp[0][0][0] = 0;
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    auto [b, a, num] = m[i];
    for (long long(j) = (long long)(0); j < (long long)(k + 1); j++) {
      for (long long(l) = (long long)(0); l < (long long)(n - k + 1); l++) {
        if (dp[i][j][l] == -INF) continue;
        if (l != n - k) chmax(dp[i + 1][j][l + 1], dp[i][j][l] + (k - 1) * b);
        if (j != k) chmax(dp[i + 1][j + 1][l], dp[i][j][l] + j * b + a);
      }
    }
  }
  cout << 2 * n - k << "\n";
  pair<long long, long long> now = {k, n - k};
  vector<long long> s, t;
  for (int i = n - 1; i >= 0; i--) {
    auto [b, a, num] = m[i];
    auto [j, l] = now;
    if (l && dp[i + 1][j][l] == dp[i][j][l - 1] + (k - 1) * b) {
      t.push_back(num);
      now.second--;
    } else {
      s.push_back(num);
      now.first--;
    }
  }
  reverse((s).begin(), (s).end());
  for (long long(i) = (long long)(0); i < (long long)(k - 1); i++)
    cout << s[i] << " ";
  for (long long(i) = (long long)(0); i < (long long)(n - k); i++)
    cout << t[i] << " " << -1 * t[i] << " ";
  cout << s.back() << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  while (n--) solve();
  return 0;
}