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;
int mp[50][50];
int main() {
  int t, n, p;
  scanf("%d", &t);
  for (int tcnt = 0; tcnt < t; tcnt++) {
    scanf("%d%d", &n, &p);
    memset(mp, 0, sizeof(mp));
    for (int i = 0; i < n; i++)
      mp[i][(i + 1) % n] = mp[(i + 1) % n][i] = mp[i][(i + 2) % n] =
          mp[(i + 2) % n][i] = 1;
    int t0 = p;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++)
        if (i - j && !mp[i][j]) {
          if (!t0) break;
          mp[i][j] = mp[j][i] = 1;
          t0--;
          if (!t0) break;
        }
      if (!t0) break;
    }
    for (int i = 0; i < n; i++)
      for (int j = i + 1; j < n; j++)
        if (mp[i][j]) printf("%d %d\n", i + 1, j + 1);
  }
  return 0;
}
