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 warning(disable : 4996)
using namespace std;
int n, p;
int mp[50][50];
int main() {
  int t, x, y, z, ca = 1;
  scanf("%d", &(t));
  while (t--) {
    scanf("%d%d", &(n), &(p));
    for (int i = (int)(1); i <= (int)(n - 2); ++i)
      mp[i][i + 1] = mp[i + 1][i] = 1, mp[i][i + 2] = mp[i + 2][i] = 1;
    mp[n - 1][n] = mp[n][n - 1] = 1, mp[n][1] = mp[1][n] = 1,
           mp[n - 1][1] = mp[1][n - 1] = 1, mp[2][n] = mp[n][2] = 1;
    int cnt = 0;
    for (int i = (int)(1); i <= (int)(n); ++i)
      for (int j = (int)(1); j <= (int)(n); ++j) {
        if (i != j && !mp[i][j]) {
          if (cnt == p) goto sym;
          cnt++;
          mp[i][j] = mp[j][i] = 1;
        }
      }
  sym:
    for (int i = (int)(1); i <= (int)(n); ++i)
      for (int j = (int)(1); j <= (int)(n); ++j) {
        if (mp[i][j]) {
          printf("%d %d\n", i, j);
          mp[i][j] = mp[j][i] = 0;
        }
      }
    memset(mp, 0, sizeof(mp));
  }
  return 0;
}
