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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
using pq = priority_queue<T>;
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
int scan() { return getchar(); }
void scan(int& a) { cin >> a; }
void scan(long long& a) { cin >> a; }
void scan(char& a) { cin >> a; }
void scan(double& a) { cin >> a; }
void scan(long double& a) { cin >> a; }
void scan(char a[]) { scanf("%s", a); }
void scan(string& a) { cin >> a; }
template <class T>
void scan(vector<T>&);
template <class T, size_t size>
void scan(array<T, size>&);
template <class T, class L>
void scan(pair<T, L>&);
template <class T, size_t size>
void scan(T (&)[size]);
template <class T>
void scan(vector<T>& a) {
  for (auto& i : a) scan(i);
}
template <class T>
void scan(deque<T>& a) {
  for (auto& i : a) scan(i);
}
template <class T, size_t size>
void scan(array<T, size>& a) {
  for (auto& i : a) scan(i);
}
template <class T, class L>
void scan(pair<T, L>& p) {
  scan(p.first);
  scan(p.second);
}
template <class T, size_t size>
void scan(T (&a)[size]) {
  for (auto& i : a) scan(i);
}
template <class T>
void scan(T& a) {
  cin >> a;
}
void IN() {}
template <class Head, class... Tail>
void IN(Head& head, Tail&... tail) {
  scan(head);
  IN(tail...);
}
string stin() {
  string s;
  cin >> s;
  return s;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> iota(int n) {
  vector<int> a(n);
  iota(begin(a), end(a), 0);
  return a;
}
template <class T>
void UNIQUE(vector<T>& x) {
  sort(begin(x), end(x));
  x.erase(unique(begin(x), end(x)), x.end());
}
int in() {
  int x;
  cin >> x;
  return x;
}
long long lin() {
  unsigned long long x;
  cin >> x;
  return x;
}
void print() { putchar(' '); }
void print(bool a) { cout << a; }
void print(int a) { cout << a; }
void print(long long a) { cout << a; }
void print(char a) { cout << a; }
void print(string& a) { cout << a; }
void print(double a) { cout << a; }
template <class T>
void print(const vector<T>&);
template <class T, size_t size>
void print(const array<T, size>&);
template <class T, class L>
void print(const pair<T, L>& p);
template <class T, size_t size>
void print(const T (&)[size]);
template <class T>
void print(const vector<T>& a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
  cout << endl;
}
template <class T>
void print(const deque<T>& a) {
  if (a.empty()) return;
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, size_t size>
void print(const array<T, size>& a) {
  print(a[0]);
  for (auto i = a.begin(); ++i != a.end();) {
    cout << " ";
    print(*i);
  }
}
template <class T, class L>
void print(const pair<T, L>& p) {
  cout << '(';
  print(p.first);
  cout << ",";
  print(p.second);
  cout << ')';
}
template <class T, size_t size>
void print(const T (&a)[size]) {
  print(a[0]);
  for (auto i = a; ++i != end(a);) {
    cout << " ";
    print(*i);
  }
}
template <class T>
void print(const T& a) {
  cout << a;
}
int out() {
  putchar('\n');
  return 0;
}
template <class T>
int out(const T& t) {
  print(t);
  putchar('\n');
  return 0;
}
template <class Head, class... Tail>
int out(const Head& head, const Tail&... tail) {
  print(head);
  putchar(' ');
  out(tail...);
  return 0;
}
long long gcd(long long a, long long b) {
  while (b) {
    long long c = b;
    b = a % b;
    a = c;
  }
  return a;
}
long long lcm(long long a, long long b) {
  if (!a || !b) return 0;
  return a * b / gcd(a, b);
}
vector<pair<long long, long long>> factor(long long x) {
  vector<pair<long long, long long>> ans;
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back({i, 1});
      while ((x /= i) % i == 0) ans.back().second++;
    }
  if (x != 1) ans.push_back({x, 1});
  return ans;
}
vector<int> divisor(int x) {
  vector<int> ans;
  for (int i = 1; i * i <= x; i++)
    if (x % i == 0) {
      ans.push_back(i);
      if (i * i != x) ans.push_back(x / i);
    }
  return ans;
}
int popcount(long long x) { return __builtin_popcountll(x); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int n) { return uniform_int_distribution<int>(0, n)(rng); }
template <class... T>
void err(const T&...) {}
template <typename Monoid>
struct SegmentTree {
  using F = function<Monoid(Monoid, Monoid)>;
  int sz;
  vector<Monoid> seg;
  const F f;
  const Monoid M1;
  SegmentTree(int n, const F f, const Monoid& M1) : f(f), M1(M1) {
    sz = 1;
    while (sz < n) sz <<= 1;
    seg.assign(2 * sz, M1);
  }
  void set(int k, const Monoid& x) { seg[k + sz] = x; }
  void build() {
    for (int k = sz - 1; k > 0; k--) {
      seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
    }
  }
  void update(int k, const Monoid& x) {
    k += sz;
    seg[k] += x;
    while (k >>= 1) {
      seg[k] = f(seg[2 * k + 0], seg[2 * k + 1]);
    }
  }
  Monoid query(int a, int b) {
    Monoid L = M1, R = M1;
    for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
      if (a & 1) L = f(L, seg[a++]);
      if (b & 1) R = f(seg[--b], R);
    }
    return f(L, R);
  }
  Monoid operator[](const int& k) const { return seg[k + sz]; }
  template <typename C>
  int find_subtree(int a, const C& check, Monoid& M, bool type) {
    while (a < sz) {
      Monoid nxt = type ? f(seg[2 * a + type], M) : f(M, seg[2 * a + type]);
      if (check(nxt))
        a = 2 * a + type;
      else
        M = nxt, a = 2 * a + 1 - type;
    }
    return a - sz;
  }
  template <typename C>
  int find_first(int a, const C& check) {
    Monoid L = M1;
    if (a <= 0) {
      if (check(f(L, seg[1]))) return find_subtree(1, check, L, false);
      return -1;
    }
    int b = sz;
    for (a += sz, b += sz; a < b; a >>= 1, b >>= 1) {
      if (a & 1) {
        Monoid nxt = f(L, seg[a]);
        if (check(nxt)) return find_subtree(a, check, L, false);
        L = nxt;
        ++a;
      }
    }
    return -1;
  }
  template <typename C>
  int find_last(int b, const C& check) {
    Monoid R = M1;
    if (b >= sz) {
      if (check(f(seg[1], R))) return find_subtree(1, check, R, true);
      return -1;
    }
    int a = sz;
    for (b += sz; a < b; a >>= 1, b >>= 1) {
      if (b & 1) {
        Monoid nxt = f(seg[--b], R);
        if (check(nxt)) return find_subtree(b, check, R, true);
        R = nxt;
      }
    }
    return -1;
  }
  void print(int n) {
    for (int i = 0; i < n; i++) {
      cerr << seg[i + sz] << " ";
    }
    cerr << '\n';
  }
};
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int T = in();
  while (T--) {
    int n = in(), k = in();
    auto id = iota(n);
    vector<long long> a, b;
    for (long long i = 0; i < n; ++i) {
      a.push_back(in()), b.push_back(in());
    }
    sort(begin(id), end(id), [&](int x, int y) { return b[x] < b[y]; });
    long long dp[100][100] = {};
    for (long long i = 0; i < n + 1; ++i)
      for (long long j = 0; j < n + 1; ++j) dp[i][j] = -1e15;
    dp[0][0] = 0;
    int pre[100][100] = {};
    int mem[100][100];
    for (long long i = 0; i < n; ++i) {
      int now = id[i];
      for (long long j = 0; j < k + 1; ++j) {
        if (chmax(dp[i + 1][j], dp[i][j] + b[now] * (k - 1))) pre[i + 1][j] = j;
        if (chmax(dp[i + 1][j + 1], dp[i][j] + a[now] + b[now] * j))
          pre[i + 1][j + 1] = j;
      }
    }
    vector<int> ope;
    int now = k;
    for (long long i = n; i >= 1; --i) {
      if (pre[i][now] != now) {
        ope.emplace_back(id[i - 1]);
        now--;
      }
    }
    reverse(begin(ope), end(ope));
    cout << n + (n - k) << '\n';
    for (long long i = 0; i < k - 1; ++i) cout << ope[i] + 1 << " ";
    vector<bool> used(n);
    for (long long i = 0; i < k; ++i) used[ope[i]] = 1;
    for (long long i = 0; i < n; ++i)
      if (!used[i]) {
        cout << i + 1 << " " << -(i + 1) << " ";
      }
    cout << ope[k - 1] + 1 << '\n';
  }
}