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 double EPS = 1e-8;
const int MAXN = 10005;
const int INF = 1111111111;
struct Node {
  int id, val;
  bool operator<(const Node& a) const {
    if (a.val == val) return a.id < id;
    return a.val < val;
  }
};
int du[100];
int n, p, tot, fg;
priority_queue<Node> q;
int main() {
  int ca;
  scanf("%d", &ca);
  while (ca--) {
    while (!q.empty()) q.pop();
    memset(du, 0, sizeof(du));
    scanf("%d%d", &n, &p);
    tot = (2 * n + p) * 2;
    fg = tot % n;
    tot /= n;
    Node ft;
    for (int i = 1; i < n; i++) {
      ft.id = i + 1;
      ft.val = 0;
      q.push(ft);
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
      while (1) {
        if (cnt == 2 * n + p) break;
        ft = q.top();
        while (ft.id <= i) {
          q.pop();
          ft = q.top();
        }
        if (fg && du[i] == tot + 1) {
          fg--;
          break;
        } else if (fg == 0 && du[i] == tot)
          break;
        q.pop();
        du[i]++, du[ft.id]++;
        ft.val++;
        q.push(ft);
        printf("%d %d\n", i, ft.id);
        cnt++;
      }
    }
  }
  return 0;
}
