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

def count(Q, k):
    size = len(Q)
    dp = [[0] * size for _ in range(k + 1)]
    for used in range(size):
        dp[0][used] = 1

    for taken in range(1, k + 1):
        for used in range(taken - 1, size):
            dp[taken][used] = (dp[taken][used - 1] + dp[taken - 1][used - 1]*len(Q[used])) % M
            
#    for d in dp:
#        print(d)
    return dp[k][size - 1]

#print("!", count([[14], [51], [17]], 3))

#exit(2)

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)

    if k == 2:
        r = C(n, k)
    else:
        r = 0
        for s in range(1, n + 1):
            Q = [[g] for g in G[s]]
            r += count(Q, k)

            V = {s}
            while len(Q) > 0:
                nQ = []
                for sq in Q:
                    nsq = []
                    for e in sq:
                        V.add(e)
                        for g in G[e]:
                            if g not in V:
                                nsq.append(g)
                    if len(nsq) > 0:
                        nQ.append(nsq)

                if len(nQ) >= k:
                    r += count(nQ, k)
                if len(nQ) <= 1:
                    break
                Q = nQ

    print(r % M)