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 MulModL(long long B, long long P, long long M) {
    long long R = 0;
    while (P > 0) {
      if ((P & 1ll) == 1) {
        R = (R + B);
        if (R >= M) R -= M;
      }
      P >>= 1ll;
      B = (B + B);
      if (B >= M) B -= M;
    }
    return R;
  }
  long long MulModD(long long B, long long P, long long M) {
    long long I = ((long double)B * (long double)P / (long double)M);
    long long R = B * P - M * I;
    R = (R % M + M) % M;
    return R;
  }
  long long BigMod(long long B, long long P, long long M) {
    long long R = 1;
    while (P > 0) {
      if (P % 2 == 1) {
        R = (R * B) % M;
      }
      P /= 2;
      B = (B * B) % M;
    }
    return R;
  }
  long long BigModML(long long B, long long P, long long M) {
    long long R = 1;
    while (P > 0) {
      if (P % 2 == 1) {
        R = MulModL(R, B, M);
      }
      P /= 2;
      B = MulModL(B, B, M);
    }
    return R;
  }
  template <class T1>
  void deb(T1 e1) {
    cerr << e1 << "\n";
  }
  template <class T1, class T2>
  void deb(T1 e1, T2 e2) {
    cerr << e1 << " " << e2 << "\n";
  }
  template <class T1, class T2, class T3>
  void deb(T1 e1, T2 e2, T3 e3) {
    cerr << e1 << " " << e2 << " " << e3 << "\n";
  }
  template <class T1, class T2, class T3, class T4>
  void deb(T1 e1, T2 e2, T3 e3, T4 e4) {
    cerr << e1 << " " << e2 << " " << e3 << " " << e4 << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5>
  void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
    cerr << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5, class T6>
  void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
    cerr << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
         << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5, class T6, class T7>
  void deb(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6, T7 e7) {
    cerr << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
         << " " << e7 << "\n";
  }
  template <class T1>
  void output(T1 e1) {
    cout << e1 << "\n";
  }
  template <class T1, class T2>
  void output(T1 e1, T2 e2) {
    cout << e1 << " " << e2 << "\n";
  }
  template <class T1, class T2, class T3>
  void output(T1 e1, T2 e2, T3 e3) {
    cout << e1 << " " << e2 << " " << e3 << "\n";
  }
  template <class T1, class T2, class T3, class T4>
  void output(T1 e1, T2 e2, T3 e3, T4 e4) {
    cout << e1 << " " << e2 << " " << e3 << " " << e4 << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5>
  void output(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5) {
    cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5, class T6>
  void output(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6) {
    cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
         << "\n";
  }
  template <class T1, class T2, class T3, class T4, class T5, class T6, class T7>
  void output(T1 e1, T2 e2, T3 e3, T4 e4, T5 e5, T6 e6, T7 e7) {
    cout << e1 << " " << e2 << " " << e3 << " " << e4 << " " << e5 << " " << e6
         << " " << e7 << "\n";
  }
  template <class T>
  T GCD(T a, T b) {
    T b1;
    while (b > 0) {
      b1 = b;
      b = a % b;
      a = b1;
    }
    return a;
  }
  template <class T>
  T LCM(T a, T b) {
    return (a * b) / GCD(a, b);
  }
  const int MOD = 1000000007;
  const int MAX = 105;
  vector<vector<int> > g;
  vector<long long> dp;
  vector<vector<int> > cnt;
  int nCr[MAX][MAX];
  void precal() {
    nCr[0][0] = 1;
    for (int i = 0; i < int(MAX); i++) nCr[i][0] = nCr[i][i] = 1;
    for (int i = 1; i < MAX; i++) {
      for (int j = 1; j < i; j++) {
        nCr[i][j] = nCr[i - 1][j - 1] + nCr[i - 1][j];
        if (nCr[i][j] >= MOD) nCr[i][j] -= MOD;
      }
    }
    return;
  }
  void dfs(int root, int u, int p, int d) {
    cnt[root][d]++;
    for (auto v : g[u]) {
      if (v == p) continue;
      dfs(root, v, u, d + 1);
    }
    return;
  }
  int solve(int n, int k) {
    if (k == 2) return nCr[n][k];
    int ret = 0;
    for (int u = 0; u < int(n); u++) {
      cnt = vector<vector<int> >(n, vector<int>(n, 0));
      cnt[u][0] = 1;
      for (auto v : g[u]) dfs(v, v, u, 1);
      for (int i = 1; i < n; i++) {
        dp = vector<long long>(k + 1, 0);
        dp[0] = 1;
        for (auto v : g[u]) {
          for (int j = k; j > 0; j--) {
            dp[j] = (dp[j - 1] * cnt[v][i] + dp[j]) % MOD;
          }
        }
        ret += dp[k];
        ret %= MOD;
      }
    }
    return ret;
  }
  int main(int argc, const char* argv[]) {
    std::ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    precal();
    int t, n, k, u, v;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      g = vector<vector<int> >(n);
      for (int i = 0; i < int(n - 1); i++) {
        cin >> u >> v;
        u--;
        v--;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      output(solve(n, k));
    }
  }
  