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>
int IntMaxVal = (int)1e20;
int IntMinVal = (int)-1e20;
long long LongMaxVal = (long long)1e20;
long long LongMinVal = (int)-1e20;
template <typename T>
struct argument_type;
template <typename T, typename U>
struct argument_type<T(U)> {};
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& s) {
  is >> s.first >> s.second;
  return is;
}
void solve() {
  int n;
  cin >> n;
  ;
  int p;
  cin >> p;
  ;
  vector<pair<int, int> > edges;
  bool hasEdge[n + 1][n + 1];
  for (int i = 1; i < n + 1; ++i)
    for (int j = 1; j < n + 1; ++j) hasEdge[i][j] = false;
  vector<int> edgeCount(n + 1, 4);
  for (int i = 1; i < n + 1; ++i) {
    int to = i + 1;
    if (to > n) to -= n;
    edges.push_back(make_pair(i, to));
    hasEdge[i][to] = hasEdge[to][i] = true;
    to = i + 2;
    if (to > n) to -= n;
    edges.push_back(make_pair(i, to));
    hasEdge[i][to] = hasEdge[to][i] = true;
  }
  while (edges.size() < 2 * n + p) {
    int x1 = 1, x2 = 1;
    while (edgeCount[x1] == n - 1) x1++;
    x2 = x1 + 1;
    while (hasEdge[x1][x2]) x2++;
    edges.push_back(make_pair(x1, x2));
    hasEdge[x1][x2] = hasEdge[x2][x1] = true;
    edgeCount[x1]++;
    edgeCount[x2]++;
  }
  for (auto p : edges) cout << p.first << ' ' << p.second << endl;
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  ;
  while (t-- > 0) solve();
}
