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;
  inline long long read() {
    int ch = getchar(), f = 1;
    long long x = 0;
    for (; ch < '0' || ch > '9'; ch = getchar())
      if (ch == '-') f = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
    return f * x;
  }
  int solve();
  int main() {
    for (int t = read(); t--; solve())
      ;
    return 0;
  }
  const int mod = 1e9 + 7;
  vector<vector<int>> e(105);
  int cnt[105];
  long long sum[105][105];
  long long fac[105], inv[105];
  long long qpow(long long x, long long k) {
    long long ans = 1;
    for (; k; k >>= 1, x = x * x % mod) {
      if (k & 1) ans = ans * x % mod;
    }
    return ans;
  }
  void dfs(int x, int fa, int dep) {
    cnt[dep]++;
    for (auto t : e[x]) {
      if (t == fa) continue;
      dfs(t, x, dep + 1);
    }
  }
  long long C(int n, int m) {
    if (n < m) return 0;
    return fac[n] * inv[m] % mod * inv[n - m] % mod;
  }
  int n, k;
  int solve() {
    n = read(), k = read();
    fac[0] = inv[0] = 1;
    for (int i = 1; i <= n; ++i)
      fac[i] = fac[i - 1] * i % mod, inv[i] = qpow(fac[i], mod - 2);
    if (k == 1) {
      for (int i = 1; i < n; ++i) read(), read();
      cout << n << endl;
      return 0;
    }
    if (k == 2) {
      for (int i = 1; i < n; ++i) read(), read();
      cout << 1ll * n * (n - 1) / 2 % mod << endl;
      return 0;
    }
    for (int i = 1; i <= n; ++i) e[i].clear();
    for (int i = 1; i < n; ++i) {
      int u = read(), v = read();
      e[u].push_back(v);
      e[v].push_back(u);
    }
    long long ans = 0;
    memset(sum, 0, sizeof sum);
    for (int i = 1; i <= n; ++i) {
      memset(sum, 0, sizeof sum);
      for (int i = 1; i <= n; ++i) sum[i][0] = 1;
      for (auto t : e[i]) {
        memset(cnt, 0, sizeof cnt);
        dfs(t, i, 1);
        for (int d = 1; d <= n; ++d) {
          for (int j = k; j >= 1; --j) {
            sum[d][j] += sum[d][j - 1] * cnt[d] % mod;
            sum[d][j] %= mod;
          }
        }
      }
      for (int d = 1; d <= n; ++d) ans += sum[d][k], ans %= mod;
    }
    cout << ans << endl;
    return 0;
  }
  