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<pair<int, int> > v;
int main() {
  int g, t, i, j, n;
  int p, e;
  cin >> t;
  while (t--) {
    v.clear();
    cin >> n >> p;
    e = 2 * n + p;
    g = 0;
    for (i = 1; i <= n; i++) {
      for (j = 1; j < i; j++) {
        if (e == 0) {
          for (int k = 0; k < v.size(); k++) {
            cout << v[k].first << " " << v[k].second << endl;
          }
          g = 1;
          break;
        }
        if (v.size() >= (2 * i + p)) {
          break;
        }
        v.push_back(make_pair(i, j));
        e--;
      }
      if (g == 1) {
        break;
      }
    }
    if (g == 1) continue;
    for (i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
  }
  return 0;
}
