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;
const int inf = 1000 * 1000 * 1000;
int mtr[30][30];
vector<int> pr;
bool prov(vector<int> v, int p) {
  int cnt = 0;
  for (int i = 0; i < v.size(); i++) {
    for (int j = i + 1; j < v.size(); j++) {
      if (mtr[v[i]][v[j]]) cnt++;
    }
  }
  return cnt > (v.size() * 2 + p);
}
bool flag = false;
void perebor(int n, int p, int v) {
  if (v == n) {
    flag |= prov(pr, p);
    return;
  }
  pr.push_back(v);
  perebor(n, p, v + 1);
  pr.pop_back();
  perebor(n, p, v + 1);
}
int main() {
  int t;
  cin >> t;
  while (t) {
    t--;
    int n, p;
    cin >> n >> p;
    memset(mtr, 0, sizeof(mtr));
    for (int i = 0; i < n; i++) {
      mtr[i][(i + 1) % n] = 1;
      mtr[(i + 1) % n][i] = 1;
    }
    if (n % 2) {
      for (int i = 0; i < n; i++) {
        mtr[(2 * i) % n][(2 * (i + 1)) % n] = 1;
        mtr[(2 * (i + 1)) % n][(2 * i) % n] = 1;
      }
    } else {
      for (int i = 0; i < n; i += 2) {
        mtr[i][(i + 2) % n] = 1;
        mtr[(i + 2) % n][i] = 1;
      }
      for (int i = 1; i < n; i += 2) {
        mtr[i][(i + 2) % n] = 1;
        mtr[(i + 2) % n][i] = 1;
      }
    }
    for (int i = 0; i < n; i++) {
      if (p == 0) break;
      for (int j = i + 1; j < n; j++) {
        if (p == 0) break;
        if (mtr[i][j] == 0) {
          mtr[i][j] = mtr[j][i] = 1;
          p--;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (mtr[i][j]) {
          cout << i + 1 << " " << j + 1 << endl;
        }
      }
    }
  }
  return 0;
}
