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 toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, 1, -1};
int t, n, p;
bool con[30][30];
int main() {
  cin >> t;
  while (t--) {
    memset((con), 0, sizeof(con));
    cin >> n >> p;
    vector<pair<int, int> > vec;
    for (int i = 1; i < n; ++i) {
      vec.push_back(make_pair(i, i + 1));
      con[i][i + 1] = con[i + 1][i] = true;
    }
    con[1][n] = con[n][1] = true;
    vec.push_back(make_pair(1, n));
    int r = n + p;
    for (int i = 1; i <= n; ++i) {
      for (int j = i + 1; j <= n; ++j) {
        if (!con[i][j]) {
          vec.push_back(make_pair(i, j));
          --r;
          if (r == 0) break;
        }
      }
      if (r == 0) break;
    }
    sort(vec.begin(), vec.end());
    for (int i = (0); i < (vec.size()); ++i)
      cout << vec[i].first << " " << vec[i].second << "\n";
  }
  return 0;
}
