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;
inline int ni() {
  int a;
  scanf("%d", &a);
  return a;
}
inline double nf() {
  double a;
  scanf("%lf", &a);
  return a;
}
template <class T>
void out(T a, T b) {
  bool first = true;
  for (T i = a; i != b; i++) {
    if (!first) printf(" ");
    first = false;
    cout << *i;
  }
  puts("");
}
template <class T>
void outl(T a, T b) {
  for (T i = a; i != b; i++) cout << *i << "\n";
}
const int MAXN = 25;
int t;
int n, p;
bool d[MAXN][MAXN];
int main() {
  t = ni();
  while (t--) {
    n = ni();
    p = ni();
    for (int i = 0; i < (int)(MAXN); i++)
      for (int j = 0; j < (int)(MAXN); j++) d[i][j] = false;
    for (int i = 0; i + 5 <= n; i += 5) {
      for (int j = i; j < i + 5; j++)
        for (int k = j + 1; k < i + 5; k++) d[j][k] = true;
    }
    int rest = n % 5;
    if (rest) {
      int v = n - rest + 1;
      for (int i = v; i <= n; i++) {
        d[0][i - 1] = d[1][i - 1] = true;
      }
    }
    for (int i = 0; i < n; i++)
      for (int j = i + 1; j < n; j++) {
        if (p && !d[i][j]) {
          p--;
          d[i][j] = true;
        }
      }
    for (int i = 0; i < (int)(n); i++) {
      for (int j = 0; j < (int)(n); j++) {
        if (d[i][j]) cout << i + 1 << " " << j + 1 << endl;
      }
    }
  }
}
