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)
    dist = [[0] * n for _ in range(n)]

    def bfs(start):
        q = deque([(0, start, -1)])
        while q:
            step, i, par = q.popleft()
            dist[start][i] = step
            for j in G[i]:
                if j == par: continue
                q.append((step + 1, j, i))

    for i in range(n): bfs(i)
    F = [0] * (n + 1)
    F[0] = 1
    for i in range(1, n + 1):
        F[i] = i * F[i - 1] % mod
    iF = [0] * (n + 1)
    iF[-1] = pow(F[-1], mod - 2, mod)
    for i in range(n - 1, -1, -1):
        iF[i] = (i + 1) * iF[i + 1] % mod

    def C(n, k):
        if n < k or k < 0: return 0
        return F[n] * iF[k] * iF[n - k] % mod

    ans = 0
    for i in range(n):
        cnt = Counter(dist[i])
        for x, v in cnt.items():
            ans += C(v, k)
            ans %= mod
    print(ans)