A tree is an undirected connected graph without cycles.

You are given a tree of n vertices. Find the number of ways to choose exactly k vertices in this tree (i. e. a k-element subset of vertices) so that all pairwise distances between the selected vertices are equal (in other words, there exists an integer c such that for all u, v (u ≠ v, u, v are in selected vertices) d_{u,v}=c, where d_{u,v} is the distance from u to v).

Since the answer may be very large, you need to output it modulo 10^9 + 7.

Input

The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.

Each test case is preceded by an empty line.

Each test case consists of several lines. The first line of the test case contains two integers n and k (2 ≤ k ≤ n ≤ 100) — the number of vertices in the tree and the number of vertices to be selected, respectively. Then n - 1 lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) which describe a pair of vertices connected by an edge. It is guaranteed that the given graph is a tree and has no loops or multiple edges.

Output

For each test case output in a separate line a single integer — the number of ways to select exactly k vertices so that for all pairs of selected vertices the distances between the vertices in the pairs are equal, modulo 10^9 + 7 (in other words, print the remainder when divided by 1000000007).

Example

Input


3

4 2
1 2
2 3
2 4

3 3
1 2
2 3

5 3
1 2
2 3
2 4
4 5


Output


6
0
1
import sys
input = sys.stdin.readline
from collections import deque, Counter

mod = 10 ** 9 + 7
for _ in range(int(input())):
    input()
    n, k = map(int, input().split())
    G = [[] for _ in range(n)]
    for _ in range(n - 1):
        a, b = map(int, input().split())
        a -= 1
        b -= 1
        G[a].append(b)
        G[b].append(a)

    if k == 2:
        print(n * (n - 1) // 2 % mod)
        continue

    ans = 0
    for r in range(n):
        cnt = [0] * n
        dp = [[0] * (k + 1) for _ in range(n)]
        for i in range(1, n):
            dp[i][0] = 1

        def dfs(i, par, d):
            cnt[d] += 1
            for j in G[i]:
                if j == par: continue
                dfs(j, i, d + 1)

        for x in G[r]:
            cnt = [0] * n
            dfs(x, r, 1)
            for i in range(1, n):
                if cnt[i] == 0: break
                for j in range(k, 0, -1):
                    dp[i][j] += dp[i][j - 1] * cnt[i]
                    dp[i][j] %= mod

        for i in range(1, n):
            ans += dp[i][k]
            ans %= mod

    print(ans)
