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


class SegmentTree:
    def __init__(self, array):
        self.n = len(array)
        self.size = 1
        while self.size < self.n:
            self.size *= 2
        self.func = lambda x, y: x if x[0] < y[0] else y
        self.default = (10 ** 9, -1)
        self.data = [self.default] * (2 * self.size)
        self.process(array)
    
    def process(self, array):
        self.data[self.size: self.size + self.n] = array
        for i in range(self.size - 1, -1, -1):
            self.data[i] = self.func(self.data[2 * i], self.data[2 * i + 1])
    
    def query(self, alpha, omega):
        """Returns the result of function over the range (inclusive)!"""
        if alpha == omega:
            return self.data[alpha + self.size]
        res = self.default
        alpha += self.size
        omega += self.size + 1
        while alpha < omega:
            if alpha & 1:
                res = self.func(res, self.data[alpha])
                alpha += 1
            if omega & 1:
                omega -= 1
                res = self.func(res, self.data[omega])
            alpha >>= 1
            omega >>= 1
        return res
    
    def update(self, index, value):
        """Updates the element at index to given value!"""
        index += self.size
        self.data[index] = value
        index >>= 1
        while index:
            self.data[index] = self.func(self.data[2 * index], self.data[2 * index + 1])
            index >>= 1


class LCA:
    def __init__(self, graph, root):
        self.graph = graph
        self.n = len(graph)
        self.euler = []
        self.first = [-1] * self.n
        self.st = None
        self.process(root)
    
    def process(self, root):
        visited, parents, heights = [False] * self.n, [-1] * self.n, [1] * self.n
        stack = [root]
        while stack:
            v = stack[-1]
            if not visited[v]:
                visited[v] = True
                self.euler += [v]
                if self.first[v] == -1:
                    self.first[v] = len(self.euler) - 1
                for u in self.graph[v]:
                    if not visited[u]:
                        stack.append(u)
                        parents[u], heights[u] = v, heights[v] + 1
            else:
                self.euler += [parents[stack.pop()]]
        self.euler = [(heights[k], k) for k in self.euler]
        self.st = SegmentTree(self.euler)
    
    def query(self, x, y):
        """Returns the lowest common ancestor of nodes x and y!"""
        p, q = min(self.first[x], self.first[y]), max(self.first[x], self.first[y])
        return self.st.query(p, q)[1]


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


def shortest_path(graph, alpha, omega):
    """Returns the shortest path between two vertices!"""
    used, dist, parents = bfs(graph, alpha)
    if not used[omega]:
        return []
    path = [omega]
    v = omega
    while parents[v] != -1:
        path += [parents[v]]
        v = parents[v]
    return path[::-1]


max_n=200
mod=10 ** 9 + 7
if 1:
    max_n = min(max_n, mod - 1)
    
    fact, inv_fact = [0] * (max_n + 1), [0] * (max_n + 1)
    fact[0] = 1
    for i in range(max_n):
        fact[i + 1] = fact[i] * (i + 1) % mod
    
    inv_fact[-1] = pow(fact[-1], mod - 2, mod)
    for i in reversed(range(max_n)):
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod
    
    def nCr_mod(n, r):
        res = 1
        while n or r:
            a, b = n % mod, r % mod
            if a < b:
                return 0
            res = res * fact[a] % mod * inv_fact[b] % mod * inv_fact[a - b] % mod
            n //= mod
            r //= mod
        return res


def sumOfProduct(arr, n, k):
    # Initialising all the values to 0
    dp = [[0 for x in range(n + 1)] for y in range(n + 1)]
    cur_sum = 0
    
    # For k = 1, the answer will simply
    # be the sum of all the elements
    for i in range(1, n + 1):
        dp[1][i] = arr[i - 1]
        cur_sum += arr[i - 1]
    
    # Filling the table in bottom up manner
    for i in range(2, k + 1):
        
        # To store the elements of the current
        # row so that we will be able to use this sum
        # for subsequent values of k
        temp_sum = 0
        
        for j in range(1, n + 1):
            # We will subtract previously computed value
            # so as to get the sum of elements from j + 1
            # to n in the (i - 1)th row
            cur_sum -= dp[i - 1][j]
            
            dp[i][j] = arr[j - 1] * cur_sum
            dp[i][j] %= mod
            temp_sum += dp[i][j]
        cur_sum = temp_sum
    return cur_sum % mod

from random import randint
    
for _ in range(int(input()) if True else 1):
    ok = 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()
    graph = [[] for _ in range(n+1)]
    for i in range(n-1):
        x, y = map(int, input().split())
        graph[x] += [y]
        graph[y] += [x]
    if k == 2:
        ans = (n * (n-1)) // 2
        print(ans % mod)
        continue
    
    root = randint(1, n)
    lca = LCA(graph, root)
    
    distances, parents = bfs(graph, root)
    
    def dist(x, y):
        return distances[x] + distances[y] - 2 * distances[lca.query(x, y)]
    ans = 0
    for i in range(1, n+1):
        count = [[0]*(n+3) for _ in range(n +1)]
        one = 0
        for j in range(1, n+1):
            if dist(i, j) == 1:one += 1
            elif dist(i, j) != 0:
                if lca.query(i, j) == i:
                    count[dist(i, j)][parents[j]] += 1
                else:
                    count[dist(i, j)][-1] += 1
        
        for x in count:
            p = []
            for y in x:
                if y:
                    p+=[y]
            if len(p) >= k:
                #print(i,p)
                ans += sumOfProduct(p, len(p), k)
                ans %= mod
                
                
                
        ans += nCr_mod(one, k)
    
    def check():
        from itertools import combinations
        ans = 0
        for p in combinations(list(range(1, n+1)), k):
            a = list(p)
            st = set()
            for x, y in combinations(a, 2):
                st.add(dist(x, y))
                if len(st) > 1:break
            if len(st) == 1:
                print("->",a)
                ans += 1
        return ans
    #print(check())
    print(ans)