Let's call an undirected graph of n vertices p-interesting, if the following conditions fulfill: 

  * the graph contains exactly 2n + p edges; 
  * the graph doesn't contain self-loops and multiple edges; 
  * for any integer k (1 ≤ k ≤ n), any subgraph consisting of k vertices contains at most 2k + p edges. 



A subgraph of a graph is some set of the graph vertices and some set of the graph edges. At that, the set of edges must meet the condition: both ends of each edge from the set must belong to the chosen set of vertices. 

Your task is to find a p-interesting graph consisting of n vertices.

Input

The first line contains a single integer t (1 ≤ t ≤ 5) — the number of tests in the input. Next t lines each contains two space-separated integers: n, p (5 ≤ n ≤ 24; p ≥ 0; <image>) — the number of vertices in the graph and the interest value for the appropriate test. 

It is guaranteed that the required graph exists.

Output

For each of the t tests print 2n + p lines containing the description of the edges of a p-interesting graph: the i-th line must contain two space-separated integers ai, bi (1 ≤ ai, bi ≤ n; ai ≠ bi) — two vertices, connected by an edge in the resulting graph. Consider the graph vertices numbered with integers from 1 to n. 

Print the answers to the tests in the order the tests occur in the input. If there are multiple solutions, you can print any of them.

Examples

Input

1
6 0


Output

1 2
1 3
1 4
1 5
1 6
2 3
2 4
2 5
2 6
3 4
3 5
3 6
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:256000000")
using namespace std;
bool isGood(vector<vector<bool> > used, int p) {
  int n = used.size();
  for (int i = 0; i < (1 << n); ++i) {
    int cnt = 0, m = 0;
    for (int j = 0; j < n; ++j) {
      if (i & (1 << j)) {
        ++cnt;
        for (int k = j + 1; k < n; ++k) {
          if ((i & (1 << k)) && used[j][k]) {
            ++m;
          }
        }
      }
    }
    if (m > cnt * 2 + p) {
      return false;
    }
  }
  return true;
}
void solve(int n, int p) {
  vector<vector<bool> > used(n, vector<bool>(n, false));
  for (int i = 0; i < n; ++i) {
    int x = i % n;
    int y = (i + 1) % n;
    int z = (i + 2) % n;
    printf("%d %d\n", x + 1, y + 1);
    printf("%d %d\n", x + 1, z + 1);
    used[x][y] = used[y][x] = true;
    used[x][z] = used[z][x] = true;
  }
  for (int i = 0; i < p; ++i) {
    while (true) {
      int u = rand() % n;
      int v = rand() % n;
      if (u == v || used[u][v]) {
        continue;
      }
      printf("%d %d\n", u + 1, v + 1);
      used[u][v] = used[v][u] = true;
      break;
    }
  }
  return;
  if (!isGood(used, p)) {
    cerr << n << " " << p << ": BAD" << endl;
    exit(0);
  } else {
    cerr << n << " " << p << ": GOOD" << endl;
  }
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) {
    int n, p;
    scanf("%d%d", &n, &p);
    solve(n, p);
  }
  return 0;
}
