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
import java.util.Scanner;

public class A {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int tc = in.nextInt();
        while (tc-- > 0) {
            int n = in.nextInt();
            int p = in.nextInt();
            boolean[][] g = new boolean[n][n];
            for (int i = 0; i < g.length; i++) {
                g[i][(i + 1) % n] = g[(i + 1) % n][i] = true;
                g[i][(i + 2) % n] = g[(i + 2) % n][i] = true;
            }
            for (int i = 0; i < n && p > 0; i++) {
                for (int j = i + 1; j < n && p > 0; j++) {
                    if (!g[i][j]) {
                        g[i][j] = g[j][i] = true;
                        p--;
                    }
                }
            }
            for (int i = 0; i < n; i++)
                for (int j = i + 1; j < n; j++)
                    if (g[i][j])
                        System.out.println((i + 1) + " " + (j + 1));
        }

    }
}
