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;
  const long long MOD = 1e9 + 7;
  const long long N = 1e2 + 10;
  vector<long long> g[N];
  void dfs(long long node, long long par, long long dd, vector<long long>& fre) {
    fre[dd]++;
    for (auto x : g[node])
      if (x != par) dfs(x, node, dd + 1, fre);
  }
  long long n, k;
  long long get(vector<long long>& arr) {
    vector<long long> cur(k + 1, 0);
    cur[0] = 1;
    for (auto x : arr) {
      for (long long i = k; i >= 1; i--) {
        cur[i] += cur[i - 1] * x;
        cur[i] %= MOD;
      }
    }
    return cur[k];
  }
  long long solve(long long x) {
    long long sz = g[x].size();
    vector<vector<long long> > go(sz, vector<long long>(n + 1, 0));
    for (long long i = 0; i < sz; i++) dfs(g[x][i], x, 1, go[i]);
    long long res = 0;
    for (long long i = 1; i <= n; i++) {
      vector<long long> arr;
      for (auto v : go)
        if (v[i] > 0) arr.push_back(v[i]);
      res += get(arr);
      res %= MOD;
    }
    return res;
  }
  void solve() {
    cin >> n >> k;
    long long u, v;
    for (long long i = 0; i < n - 1; i++) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    long long res = 0;
    if (k == 2) {
      res = (n * (n - 1)) / 2;
      res %= MOD;
    } else {
      for (long long i = 1; i <= n; i++) {
        res += solve(i);
        res %= MOD;
      }
    }
    cout << res << '\n';
    for (long long i = 1; i <= n; i++) g[i].clear();
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) solve();
  }
  