description: 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
solution: #include <bits/stdc++.h>
  using namespace std;
  long long cil(long long x, long long y) {
    if (x % y == 0) {
      return x / y;
    };
    return x / y + 1;
  }
  const int32_t mod = 1e9 + 7;
  long long power(long long a, long long b) {
    long long ans = 1;
    a %= mod;
    while (b > 0) {
      if (b % 2) {
        ans = (ans * a) % mod;
      }
      a = (a * a) % mod;
      b /= 2;
    }
    return ans;
  }
  long long modi(long long a) { return power(a, mod - 2); }
  const int32_t MXN = 1e5 + 5;
  long long factorial[MXN + 1];
  long long make_factorial() {
    factorial[0] = 1;
    for (long long i = 1; i <= MXN; i++) {
      factorial[i] = i * factorial[i - 1];
      factorial[i] %= mod;
    }
    return 0;
  }
  long long ncr(long long n, long long r) {
    if (n < r) return 0;
    long long ans = factorial[n];
    ans %= mod;
    ans *= modi(factorial[r]);
    ans %= mod;
    ans *= modi(factorial[n - r]);
    ans %= mod;
    return ans;
  }
  const int32_t N = 101;
  long long n, k;
  vector<long long> adj[N];
  long long sol(vector<long long>& a) {
    if ((long long)(a).size() < k) return 0;
    long long m = (long long)(a).size();
    vector<vector<long long> > dp(m + 1, vector<long long>(k + 1, 0));
    for (long long i = 0; i <= m; i++) dp[i][0] = 1;
    for (long long i = 1; i <= m; i++) {
      for (long long j = 1; j <= i; j++) {
        if (j > k) break;
        dp[i][j] = dp[i - 1][j - 1] * a[i - 1] + dp[i - 1][j];
        dp[i][j] %= mod;
      }
    }
    return dp[m][k];
  }
  long long dfs(long long v, long long pre, long long curr, long long tar) {
    if (curr == tar) return 1;
    long long cnt = 0;
    for (auto u : adj[v]) {
      if (u == pre) continue;
      long long d = dfs(u, v, curr + 1, tar);
      cnt += d;
    }
    return cnt;
  }
  long long calc(long long root) {
    vector<long long> a[n + 1];
    for (long long i = 1; i <= n; i++) {
      for (auto u : adj[root]) {
        long long x = dfs(u, root, 0, i - 1);
        a[i].push_back(x);
      }
    }
    long long ans = 0;
    for (long long d = 1; d <= n; d++) {
      ans += sol(a[d]);
      ans %= mod;
    }
    return ans;
  }
  void solve() {
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) {
      adj[i].clear();
    }
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    long long ans = 0;
    if (k == 2) {
      ans = ncr(n, 2);
      cout << ans << '\n';
      return;
    }
    for (long long i = 1; i <= n; i++) {
      ans += calc(i);
      ans %= mod;
    }
    cout << ans << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    make_factorial();
    long long tt = 1;
    cin >> tt;
    for (long long tc = 1; tc <= tt; tc++) {
      solve();
    }
    return 0;
  }
  