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
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0
    
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# ------------------- fast io --------------------
from math import ceil


def prod(a, mod=10 ** 9 + 7):
    ans = 1
    for each in a:
        ans = (ans * each) % mod
    return ans


def gcd(x, y):
    while y:
        x, y = y, x % y
    return x


def lcm(a, b): return a * b // gcd(a, b)


def binary(x, length=16):
    y = bin(x)[2:]
    return y if len(y) >= length else "0" * (length - len(y)) + y


from collections import deque

def bfs(graph, alpha=0):
    """Breadth first search on a graph!"""
    n = len(graph)
    q = deque([alpha])
    used = [False] * n
    used[alpha] = True
    dist, parents = [0] * n, [-1] * n
    while q:
        v = q.popleft()
        for u in graph[v]:
            if not used[u]:
                used[u] = True
                q.append(u)
                dist[u] = dist[v] + 1
                parents[u] = v
    return dist, parents

for _ in range(int(input()) if True else 1):
    empty = input()
    n, k = map(int, input().split())
    # a, b = map(int, input().split())
    # c, d = map(int, input().split())
    # a = list(map(int, input().split()))
    # b = list(map(int, input().split()))
    # s = input()
    mod = 10**9 + 7
    
    graph = [[] for i in range(n + 1)]
    for __ in range(n-1):
        x, y = map(int, input().split())
        graph[x] += [y]
        graph[y] += [x]

    if k == 2:
        print((n * (n - 1)) // 2)
        continue
        
    ans = 0
    for root in range(1, n + 1):
        dist, parents = bfs(graph, root)
        check = [[0]*(n + 1) for p in range(n+1)]
        for i in range(1, n + 1):
            if i == root: continue
            x = i
            while parents[x] != root:
                x = parents[x]
            check[dist[i]][x] += 1
        
        for b in check:
            a = [x for x in b if x]
            if len(a) < k: continue
            # dp[i] = sum of every i elements
            dp = [0] * (k + 1)
            dp[0] = 1
            for i in range(len(a)):
                dp2 = [0] * (k + 1)
                for j in range(1, k + 1):
                    dp2[j] = (dp[j] + dp[j-1] * a[i]) % mod
                dp = list(dp2)
            ans = (ans + dp[-1]) % mod
    
    print(ans)