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>
#pragma comment(linker, "/STACK:268435456")
using namespace std;
const double PI = 3.1415926535897932384626433832795;
template <class T>
T min(T &a, T &b) {
  return (a < b) ? a : b;
}
template <class T>
T max(T &a, T &b) {
  return (a > b) ? a : b;
}
template <class T>
T sqr(T &a) {
  return a * a;
}
template <class T>
T abs(T &a) {
  return (a < 0) ? (-a) : a;
}
inline void read(int &x) { scanf("%d", &x); }
inline void readln(int &x) { scanf("%d\n", &x); }
inline void write(int x) { printf("%d", x); }
inline void writeln(int x) { printf("%d\n", x); }
inline void read(long long &x) { scanf("%lld", &x); }
inline void readln(long long &x) { scanf("%lld\n", &x); }
inline void write(long long x) { printf("%lld", x); }
inline void writeln(long long x) { printf("%lld\n", x); }
inline void read(double &x) { scanf("%lf", &x); }
inline void readln(double &x) { scanf("%lf\n", &x); }
inline void write(double x) { printf("%.15lf", x); }
inline void writeln(double x) { printf("%.15lf\n", x); }
inline void read(char *s) { scanf("%s", s); }
inline void readln(char *s) { scanf("%s\n", s); }
inline void write(char *s) { printf("%s", s); }
inline void writeln(char *s) { printf("%s\n", s); }
inline int readInt() {
  int x;
  read(x);
  return x;
}
inline long long readLong() {
  long long x;
  read(x);
  return x;
}
inline double readDouble() {
  double x;
  read(x);
  return x;
}
void showTime() {}
void prepareInputOutput() {}
bool a[100][100];
void solve() {
  int n, p;
  cin >> n >> p;
  memset((a), 0, sizeof(a));
  vector<int> deg(n);
  for (int i = 0; i < ((int)(2 * n + p)); i++) {
    int minV = -1;
    for (int i = 0; i < ((int)(n)); i++) {
      if (minV == -1 || deg[i] < deg[minV]) {
        minV = i;
      }
    }
    int minU = -1;
    for (int i = 0; i < ((int)(n)); i++) {
      if (i != minV && !a[i][minV] && (minU == -1 || deg[i] < deg[minU])) {
        minU = i;
      }
    }
    a[minV][minU] = true;
    a[minU][minV] = true;
    deg[minU]++;
    deg[minV]++;
    cout << minU + 1 << " " << minV + 1 << "\n";
  }
}
int main() {
  prepareInputOutput();
  int T;
  cin >> T;
  for (int _ = 0; _ < ((int)(T)); _++) solve();
  showTime();
}
