#include <bits/stdc++.h>
using namespace std;
int main() {
  int nb_cas;
  cin >> nb_cas;
  while (nb_cas--) {
    int n, p;
    cin >> n >> p;
    vector<pair<int, int> > edges;
    vector<pair<int, int> > degres(n);
    vector<vector<bool> > matrice_adjacence(n, vector<bool>(n, false));
    for (int c = 0; c < n; c++) degres[c] = make_pair(0, c);
    sort(degres.begin(), degres.end());
    for (int c = 0; c < 2 * n + p; c++) {
      int c2, c3;
      for (c2 = 0; c2 < degres.size(); c2++)
        for (c3 = 0; c3 < c2; c3++)
          if (!matrice_adjacence[degres[c2].second][degres[c3].second])
            goto end;
    end:
      int from = degres[c3].second;
      int to = degres[c2].second;
      if (from > to) swap(from, to);
      matrice_adjacence[from][to] = matrice_adjacence[to][from] = true;
      edges.push_back(make_pair(from, to));
      degres[0].first++;
      degres[1].first++;
      sort(degres.begin(), degres.end());
    }
    sort(edges.begin(), edges.end());
    for (int c = 0; c < edges.size(); c++)
      cout << edges[c].first + 1 << " " << edges[c].second + 1 << endl;
  }
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int nb_cas;
  cin >> nb_cas;
  while (nb_cas--) {
    int n, p;
    cin >> n >> p;
    vector<pair<int, int> > edges;
    vector<pair<int, int> > degres(n);
    vector<vector<bool> > matrice_adjacence(n, vector<bool>(n, false));
    for (int c = 0; c < n; c++) degres[c] = make_pair(0, c);
    sort(degres.begin(), degres.end());
    for (int c = 0; c < 2 * n + p; c++) {
      int c2, c3;
      for (c2 = 0; c2 < degres.size(); c2++)
        for (c3 = 0; c3 < c2; c3++)
          if (!matrice_adjacence[degres[c2].second][degres[c3].second])
            goto end;
    end:
      int from = degres[c3].second;
      int to = degres[c2].second;
      if (from > to) swap(from, to);
      matrice_adjacence[from][to] = matrice_adjacence[to][from] = true;
      edges.push_back(make_pair(from, to));
      degres[0].first++;
      degres[1].first++;
      sort(degres.begin(), degres.end());
    }
    sort(edges.begin(), edges.end());
    for (int c = 0; c < edges.size(); c++)
      cout << edges[c].first + 1 << " " << edges[c].second + 1 << endl;
  }
}

```