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;
long long LLMAX = 9223372036854775807LL;
const int MOD = 1000000007;
const int maxn = 111 + 10;
int n, t, p;
bool G[25][25];
int main() {
  scanf("%d", &(t));
  while (t--) {
    memset(G, 0, sizeof(G));
    scanf("%d %d", &(n), &(p));
    for (int i = 0; i < n; ++i) {
      for (int j = 1; j < 3; ++j) {
        G[i][(i + j) % n] = 1;
        G[(i + j) % n][i] = 1;
        G[i][(i - j + n) % n] = 1;
        G[(i - j + n) % n][i] = 1;
      }
    }
    int i = 0, j = 0;
    while (p > 0) {
      while (G[i][j] || i == j) {
        j++;
        if (j == n) {
          i++;
          j = 0;
        }
      }
      G[i][j] = G[j][i] = 1;
      p--;
    }
    int cc = 0;
    for (int i = 0; i < n; ++i)
      for (int j = i + 1; j < n; ++j)
        if (G[i][j]) {
          printf("%d %d\n", (i + 1), (j + 1));
          cc++;
        }
  }
}
