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 Get() {
  int res = 0, q = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') q = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res * q;
}
const double eps = 1e-12, pi = M_PI;
const int oo = (int)2e9, mod = (int)1e9 + 7;
const long long INF = (long long)1e17;
bool v[27][27];
int n, p, cnt;
int d[27];
int main() {
  int test;
  for (scanf("%d"
             "\n",
             &test);
       test--;) {
    memset(v, 0, sizeof(v)), cnt = 0;
    scanf(
        "%d"
        "%d"
        "\n",
        &n, &p);
    int m = 2 * n + p;
    for (int i = (1), end = (n); i <= end; ++i) {
      if (i <= 5) {
        for (int j = (max(1, i - 4)), end = (i - 1); j <= end; ++j)
          if (cnt < m)
            ++cnt, v[i][j] = v[j][i] = 1,
                   printf(
                       "%d"
                       " "
                       "%d"
                       "\n",
                       i, j);
      } else
        for (int j = (i - 2), end = (i - 1); j <= end; ++j)
          if (cnt < m)
            ++cnt, v[i][j] = v[j][i] = 1,
                   printf(
                       "%d"
                       " "
                       "%d"
                       "\n",
                       i, j);
      if (cnt == m) break;
    }
    if (cnt < m) {
      for (int i = (1), end = (n); i <= end; ++i)
        for (int j = (1), end = (n); j <= end; ++j) {
          if (!v[i][j] && cnt < m && i != j)
            v[i][j] = v[j][i] = 1,
            printf(
                "%d"
                " "
                "%d"
                "\n",
                i, j),
            ++cnt;
        }
    }
  }
  return 0;
}
