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>
#pragma warning(disable : 4996)
using namespace std;
const int N = 30;
int n, p, mp[N][N];
set<pair<int, int> > myset;
int main() {
  int cas;
  scanf("%d", &cas);
  while (cas--) {
    scanf("%d%d", &n, &p);
    int m = 2 * n + p;
    int used = 0;
    memset(mp, false, sizeof(mp));
    myset.clear();
    for (int i = 0; i < n; i++) myset.insert(make_pair(0, i));
    for (int i = used; i < m; i++) {
      vector<pair<int, int> > rec;
      pair<int, int> u = *myset.begin();
      myset.erase(myset.begin());
      rec.push_back(make_pair(u.first + 1, u.second));
      int del = 1;
      while (true) {
        pair<int, int> v = *myset.begin();
        myset.erase(myset.begin());
        if (mp[u.second][v.second]) {
          rec.push_back(v);
          del++;
        } else {
          del++;
          mp[u.second][v.second] = mp[v.second][u.second] = 1;
          rec.push_back(make_pair(v.first + 1, v.second));
          break;
        }
      }
      for (int j = 0; j < (int)rec.size(); j++) {
        myset.insert(rec[j]);
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (mp[i][j]) printf("%d %d\n", i + 1, j + 1);
      }
    }
  }
  return 0;
}
