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;
vector<pair<int, int> > E;
int n, p;
bool adj[25][25];
void addEdge(int u, int v) {
  if (int(E.size()) < 2 * n + p) {
    if (!adj[u][v])
      adj[u][v] = adj[v][u] = true, E.push_back(pair<int, int>(u, v));
  }
}
int main() {
  int nT;
  cin >> nT;
  while (nT--) {
    cin >> n >> p;
    E.clear();
    memset(adj, false, sizeof adj);
    for (int u = 1; u <= n; u++) {
      int v1 = u + 1, v2 = u + 2;
      if (v1 > n) v1 -= n;
      if (v2 > n) v2 -= n;
      addEdge(u, v1);
      addEdge(u, v2);
    }
    for (int u = 1; u <= n && int(E.size()) < 2 * n + p; u++)
      for (int v = u + 1; v <= n && int(E.size()) < 2 * n + p; v++) {
        if (adj[u][v]) continue;
        addEdge(u, v);
      }
    for (int i = 0; i < int(E.size()); i++)
      cout << E[i].first << ' ' << E[i].second << '\n';
  }
  return 0;
}
