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;
bool haveEdge[24][24];
void cleanEdges() {
  int i, j;
  for (i = 0; i < 24; i++) {
    for (j = 0; j < 24; j++) {
      haveEdge[i][j] = false;
    }
  }
}
int main() {
  int t;
  int n, p;
  int i, j;
  cin >> t;
  for (int t_ = 0; t_ < t; t_++) {
    cin >> n >> p;
    cleanEdges();
    for (i = 0; i < n; i++) {
      cout << (i + 1) << " " << ((i + 1) % n + 1) << endl;
      cout << (i + 1) << " " << ((i + 2) % n + 1) << endl;
      haveEdge[i][i] = true;
      haveEdge[i][(i + 1) % n] = true;
      haveEdge[i][(i + 2) % n] = true;
      haveEdge[(i + 1) % n][i] = true;
      haveEdge[(i + 2) % n][i] = true;
    }
    i = 0;
    j = 0;
    for (int p_ = 0; p_ < p; p_++) {
      while (haveEdge[i][j]) {
        j++;
        if (j == n) {
          j = 0;
          i++;
        }
      }
      cout << (i + 1) << " " << (j + 1) << endl;
      haveEdge[i][j] = true;
      haveEdge[j][i] = true;
    }
    cout << endl;
  }
  return 0;
}
