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;
  void solve();
  void _print(long long x) { cerr << x; }
  template <class T>
  void _print(vector<T> v) {
    cout << " [ ";
    for (T i : v) {
      _print(i);
      cout << " ";
    }
    cout << "]";
  }
  long long o = 0;
  long long M = 1e9 + 7;
  unsigned long long fac[101];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    chrono::high_resolution_clock::time_point t1 =
        chrono::high_resolution_clock::now();
    long long t;
    fac[0] = 1;
    for (int i = 1; i <= 100; i++) fac[i] = (fac[i - 1] * i) % M;
    cin >> t;
    while (t--) {
      solve();
      cout << endl;
    }
    chrono::high_resolution_clock::time_point t2 =
        chrono::high_resolution_clock::now();
    chrono::duration<double, std::milli> time_span = t2 - t1;
    return 0;
  }
  unsigned long long power(unsigned long long x, int y, int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  unsigned long long modInverse(unsigned long long n, int p) {
    return power(n, p - 2, p);
  }
  unsigned long long nCrModPFermat(unsigned long long n, int r, int p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  long long k;
  long long fooo(int* a, long long n) {
    vector<long long> g(n + 1), t(n + 1);
    g[0] = 1;
    for (long long d = (0); d < (n); d++) {
      for (long long i = (1); i < (n + 1); i++)
        t[i] = (-a[d] * g[i] + g[i - 1]) % M;
      t[0] = (-a[d] * g[0]) % M;
      g = t;
    }
    return abs(g[n - k]);
  }
  void solve() {
    long long n;
    cin >> n >> k;
    vector<vector<long long>> edges(n + 1);
    for (long long i = (0); i < (n - 1); i++) {
      long long x, y;
      cin >> x >> y;
      edges[x].push_back(y);
      edges[y].push_back(x);
    }
    if (k == 2) {
      cout << nCrModPFermat(n, 2, M);
      return;
    }
    vector<long long> deg(n + 1);
    map<long long, map<long long, long long>> temp;
    long long ans = 0;
    function<void(long long, long long, long long, long long)> dfs =
        [&](long long a, long long p, long long d, long long s) {
          temp[d][s]++;
          for (auto x : edges[a]) {
            if (x == p) continue;
            if (s == -1)
              dfs(x, a, d + 1, x);
            else
              dfs(x, a, d + 1, s);
          }
          return;
        };
    for (long long i = (1); i < (n + 1); i++) {
      deg[i] = edges[i].size();
    }
    for (long long i = (1); i < (n + 1); i++) {
      if (deg[i] >= k) {
        temp.clear();
        dfs(i, -1, 0, -1);
        long long p = 0;
        for (auto x : temp) {
          long long t = 1;
          int gg[101] = {0};
          int pp = 0;
          for (auto y : x.second) {
            gg[pp++] = (y.second);
          }
          if (pp < k) continue;
          ans += fooo(gg, pp);
          ans %= M;
        }
      }
    }
    cout << ans;
  }
  