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>
using namespace std;
void solve() {
  int n, p;
  int i, j, k;
  int deg[25];
  int adjmat[25][25];
  for (i = 0; i < 25; i++) {
    for (j = 0; j < 25; j++) {
      adjmat[i][j] = 0;
    }
  }
  cin >> n >> p;
  for (i = 0; i < n; i++) {
    cout << i + 1 << " " << (i + 1) % n + 1 << endl;
    adjmat[i][(i + 1) % n] = 1;
    adjmat[(i + 1) % n][i] = 1;
    cout << i + 1 << " " << (i + 2) % n + 1 << endl;
    adjmat[i][(i + 2) % n] = 1;
    adjmat[(i + 2) % n][i] = 1;
  }
  for (i = 0; i < n; i++) {
    deg[i] = 2;
  }
  for (k = 0; k < p; k++) {
    int best = 0x3FFFFFFF;
    int bi, bj;
    for (i = 0; i < n; i++) {
      for (j = i + 1; j < n; j++) {
        if (adjmat[i][j] == 0) {
          if (deg[i] + deg[j] < best) {
            bi = i;
            bj = j;
          }
        }
      }
    }
    adjmat[bi][bj] = 1;
    adjmat[bj][bi] = 1;
    cout << bi + 1 << " " << bj + 1 << endl;
    deg[bi]++;
    deg[bj]++;
  }
}
int main() {
  int n;
  int i;
  cin >> n;
  for (i = 0; i < n; i++) {
    solve();
  }
  return 0;
}
