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;
int nextint() {
  int _;
  scanf("%d", &_);
  return _;
}
long long nextll() {
  long long _;
  scanf("%I64d", &_);
  return _;
}
string nextstring() {
  string _;
  cin >> _;
  return _;
}
vector<int> sz;
int g[32][32];
class vertex {
 public:
  vertex(int id = 0) : id(id) {}
  int id;
};
bool operator<(const vertex &a, const vertex &b) {
  if (sz[a.id] != sz[b.id]) return sz[a.id] < sz[b.id];
  return a.id < b.id;
}
int main() {
  int t = nextint();
  while (t--) {
    int n = nextint();
    int p = nextint();
    sz = vector<int>(n);
    memset(g, 0, sizeof g);
    set<vertex> s;
    for (int i = 0; i < n; i++) {
      s.insert(vertex(i));
      g[i][i] = 1;
      sz[i]++;
    }
    int cnt = 2 * n + p;
    while (cnt--) {
      int i = s.begin()->id, j = -1;
      s.erase(s.begin());
      for (set<vertex>::iterator it = s.begin(); it != s.end(); it++) {
        if (g[i][it->id] == 0) {
          j = it->id;
          s.erase(it);
          break;
        }
      }
      g[i][j] = g[j][i] = 1;
      sz[i]++;
      sz[j]++;
      s.insert(vertex(i));
      s.insert(vertex(j));
    }
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (g[i][j]) {
          printf("%d %d\n", i + 1, j + 1);
        }
      }
    }
  }
  return 0;
}
