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 gra[26][26];
void intialise() {
  for (int i = 0; i <= 25; i++) {
    for (int j = 0; j <= 25; ++j) {
      if (i == 0 || j == 0 || i == j)
        gra[i][j] = 0;
      else if (i <= 5 && j <= 5)
        gra[i][j] = 1;
      else
        gra[i][j] = 0;
    }
  }
}
void add(int num, int i) {
  for (int j = 1; j <= num; j++) {
    gra[i][j] = 1;
    gra[j][i] = 1;
  }
}
int main() {
  int test;
  cin >> test;
  int n, p;
  while (test--) {
    intialise();
    cin >> n >> p;
    int temp = p;
    bool done = false;
    for (int i = 6; i <= n; i++) {
      if (done)
        add(2, i);
      else if (temp <= (((i) * (i - 5)) / 2)) {
        done = true;
        add(p + 2, i);
      } else {
        add(i - 1, i);
        p = temp - (((i) * (i - 5)) / 2);
      }
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j < i; ++j) {
        if (gra[i][j] == 1) printf("%d %d\n", i, j);
      }
    }
  }
}
