human
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
The above is tricky. Write me a correct solution in PYTHON3.
 ... Write me a Python code to find the number of ways to choose exactly k vertices in this tree (i?
assistant
#Here is Python code to find the number of ways to choose exactly k vertices in this tree (i.

import sys
input = sys.stdin.readline
from collections import defaultdict as dd

MOD = 10 ** 9 + 7

def main():
    input()
    n, k = map(int, input().split())
    edges = [[] for _ in range(n)]
    for _ in range(n - 1):
        u, v = map(int, input().split())
        u -= 1
        v -= 1
        edges[u].append(v)
        edges[v].append(u)
    if k == 2:
        print(n * (n - 1) // 2)
        return
    
    global ans
    ans = 0
    def dfs(pos, bpos):
        ret = dd(int)
        ret[0] = 1
        if bpos == -1:
            lst = dd(list)
        for npos in edges[pos]:
            if npos == bpos:
                continue
            x = dfs(npos, pos)
            for k2, v in x.items():
                k2 += 1
                ret[k2] += v
                if bpos == -1:
                    lst[k2].append(v)
        if bpos == -1:
            return lst
        else:
            return ret
    for i in range(n):
        lst = dfs(i, -1)
        for k2, v in lst.items():
            if len(v) < k:
                continue
            dp = [0] * (k + 1)
            dp[0] = 1
            for i in v:
                for j in range(k, 0, -1):
                    dp[j] += dp[j - 1] * i
                    dp[j] %= MOD
            ans += dp[k]
            ans %= MOD
    print(ans % MOD)
    
for _ in range(int(input())):
    main()