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;
void fast() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
}
void online_judge() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
const int flag_max = 0x3f3f3f3f;
const long long OO = 1e9;
const double EPS = (1e-7);
int dcmp(double x, double y) { return fabs(x - y) <= EPS ? 0 : x < y ? -1 : 1; }
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int n, p;
int F = 0;
map<pair<int, int>, bool> check;
vector<pair<int, int>> pa_ir;
void dfs(int star) {
  for (int i = star + 1; i <= n; i++) {
    if (!check[{star, i}]) {
      F++;
      check[{star, i}] = 1;
      check[{i, star}] = 1;
      pa_ir.push_back({star, i});
      dfs(i);
    }
  }
}
int main() {
  fast();
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> p;
    p = 2 * n + p;
    pa_ir.clear();
    check.clear();
    dfs(1);
    sort(((pa_ir).begin()), ((pa_ir).end()));
    for (int i = 0; i < p; i++) {
      cout << pa_ir[i].first << ' ' << pa_ir[i].second << '\n';
    }
  }
  return 0;
}
