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 inv = 1000000000;
const int minv = -inv;
const int max_n = 24 + 5;
int T;
int n, p, c;
set<pair<int, int> > S;
void remove(int u, int v) {
  if (u > v) swap(u, v);
  if (S.find(pair<int, int>(u, v)) != S.end()) {
    printf("%d ", u + 1);
    printf("%d\n", v + 1);
    S.erase(pair<int, int>(u, v));
    --c;
  }
};
int main() {
  scanf("%d", &T);
  for (int z = 0; z < T; ++z) {
    scanf("%d", &n);
    scanf("%d", &p);
    c = 2 * n + p;
    S.clear();
    for (int j = 0; j < n; ++j)
      for (int i = 0; i < j; ++i) S.insert(pair<int, int>(i, j));
    for (int i = 0; i < n; ++i) {
      remove(i, (i + 1) % n);
      remove(i, (i + 2) % n);
      remove(i, (i + n - 1) % n);
      remove(i, (i + n - 2) % n);
    }
    assert(c == p);
    for (int j = 0; j < n; ++j)
      for (int i = 0; i < j; ++i) {
        if (c == 0)
          goto labelafterloops;
        else
          remove(i, j);
      }
  labelafterloops:;
  }
  return 0;
}
