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;
vector<int> num[50];
bool g[50][50];
int main() {
  int t, n, p;
  cin >> t;
  for (int i = 0; i < t; i++) {
    cin >> n >> p;
    for (int j = 0; j < 50; j++) {
      num[j].clear();
    }
    for (int j = 0; j < 50; j++) {
      for (int k = 0; k < 50; k++) {
        g[j][k] = false;
      }
    }
    for (int i = 0; i < 5; i++) {
      for (int j = i + 1; j < 5; j++) {
        num[i].push_back(j);
        num[j].push_back(i);
        g[i][j] = true;
        g[j][i] = true;
      }
    }
    for (int j = 5; j < n; j++) {
      num[j].push_back(0);
      num[0].push_back(j);
      g[0][j] = true;
      g[j][0] = true;
      num[j].push_back(j - 1);
      num[j - 1].push_back(j);
      g[j][j - 1] = true;
      g[j - 1][j] = true;
    }
    for (int j = 0; j < p; j++) {
      bool f = true;
      for (int k = 0; k < n && f; k++) {
        for (int l = k + 1; l < n && f; l++) {
          if (!g[l][k]) {
            num[l].push_back(k);
            num[k].push_back(l);
            g[k][l] = true;
            g[l][k] = true;
            f = false;
          }
        }
      }
    }
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < num[j].size(); k++) {
        if (num[j][k] > j) {
          cout << (j + 1) << " " << num[j][k] + 1 << endl;
        }
      }
    }
  }
}
