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;
const int inf = ((1 << 30) - 1);
const long long linf = ((1ll << 62) - 1);
const int N = 100;
int n, p;
int a[N][N];
vector<pair<int, int> > ans;
void addEdge(int u, int v) {
  assert(!a[u][v] && !a[v][u]);
  ans.push_back(make_pair(u, v));
  a[u][v] = a[v][u] = 1;
}
bool check() {
  if (((int)(ans).size()) != 2 * n + p) return false;
  for (int i = 0; i < (1 << n); i++) {
    int maxCount = __builtin_popcount(i) + p;
    int curCount = 0;
    for (int j = 0; j < ((int)(ans).size()); j++)
      if ((i & (1 << ans[j].first)) && (i & (1 << ans[j].second))) curCount++;
    if (curCount > maxCount) return false;
  }
  return true;
}
void solve() {
  cin >> n >> p;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) a[i][j] = 0;
  ans.clear();
  for (int t = 1; t <= 2; t++)
    for (int i = 0; i < n; i++) addEdge(i, (i + t) % n);
  int np = p;
  for (int t = 3;; t++) {
    for (int i = 0; i < n; i++)
      if (np > 0) addEdge(i, (i + t) % n), np--;
    if (!np) break;
  }
  for (int i = 0; i < ((int)(ans).size()); i++)
    printf("%d %d\n", ans[i].first + 1, ans[i].second + 1);
}
int main() {
  int t;
  cin >> t;
  while (t--) solve(), ({});
  return 0;
}
