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 oo = (int)1e9;
const double PI = 2 * acos(0.0);
const double eps = 1e-9;
int t, n, p;
bool graph[25][25];
void build(int step) {
  bool visited[25] = {0};
  while (p) {
    int st = -1;
    for (int i = 0; i < (int)(n); i++)
      if (!visited[i]) {
        st = i;
        break;
      }
    if (st == -1) break;
    for (int i = st; !visited[i] && p; --p, graph[i][(i + step) % n] = true,
             visited[i] = true, i = (i + (1 << step)) % n)
      ;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n >> p;
    p = 2 * n + p;
    memset(graph, 0, sizeof(graph));
    for (int i = n - 1; i >= 1; --i) {
      build(i);
    }
    for (int i = 0; i < (int)(n); i++)
      for (int k = 0; k < (int)(n); k++)
        if (graph[i][k]) cout << i + 1 << " " << k + 1 << endl;
  }
  return 0;
}
