#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
void solve() {
  int n, p;
  cin >> n >> p;
  int adj[n][n];
  memset(adj, 0, sizeof(adj));
  int orig = 2 * n;
  for (ll i = 0; i < (n); i++) {
    if (orig == 0) break;
    for (int j = i + 1; j < n; j++) {
      if (orig == 0) break;
      adj[i][j] = 1;
      adj[j][i] = 1;
      orig--;
    }
  }
  for (ll i = 0; i < (n); i++) {
    if (p == 0) break;
    for (int j = i + 1; j < n; j++) {
      if (p == 0) break;
      if (adj[i][j] == 0) {
        p--;
        adj[i][j] = 1;
        adj[j][i] = 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (adj[i][j] == 1) {
        cout << i + 1 << " " << j + 1 << '\n';
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vd = vector<double>;
using vs = vector<string>;
using vpii = vector<pair<int, int>>;
using vpll = vector<pair<ll, ll>>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
int dx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
void solve() {
  int n, p;
  cin >> n >> p;
  int adj[n][n];
  memset(adj, 0, sizeof(adj));
  int orig = 2 * n;
  for (ll i = 0; i < (n); i++) {
    if (orig == 0) break;
    for (int j = i + 1; j < n; j++) {
      if (orig == 0) break;
      adj[i][j] = 1;
      adj[j][i] = 1;
      orig--;
    }
  }
  for (ll i = 0; i < (n); i++) {
    if (p == 0) break;
    for (int j = i + 1; j < n; j++) {
      if (p == 0) break;
      if (adj[i][j] == 0) {
        p--;
        adj[i][j] = 1;
        adj[j][i] = 1;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (adj[i][j] == 1) {
        cout << i + 1 << " " << j + 1 << '\n';
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```