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
from collections import defaultdict

def memoize(function):
    cache = {}
    def decorated_function(*args):
        try:
            return cache[args]
        except KeyError:
            val = function(*args)
            cache[args] = val
            return val
    return decorated_function

@memoize
def inverseMod(n, mod):
    if 1 == n: return 1

    return ((mod - mod//n) * inverseMod(mod % n, mod)) % mod

M = 10**9 + 7

F = [1, 1]
for n in range(2, 10**5 + 1):
    F.append((F[-1] * n) % M)

RF = [1, 1]
for n in range(2, 10**5 + 1):
    RF.append((RF[-1] * inverseMod(n, M)) % M)

def C(n, k):
    return (F[n] * RF[n - k] * RF[k]) % M

T = int(input())

for _ in range(T):
    input()
    n, k = map(int, input().split(' '))

    G = defaultdict(list)
    for _ in range(n - 1):
        s, e = map(int, input().split(' '))
        G[s].append(e)
        G[e].append(s)

    r = 0
    for s in range(1, n + 1):
        Q = [s]
        V = set()
        while len(Q) > 0:
            nQ = []
            for q in Q:
                V.add(q)
                for e in G[q]:
                    if e not in V:
                        nQ.append(e)
            if len(nQ) >= k:
                r += C(len(nQ), k)
            Q = nQ

    print(r % M)