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 N = 2e2 + 10;
  const long long mod = 1e9 + 7;
  const double eps = 1e-9;
  const long long INF = 0x3f3f3f3f;
  long long n, k;
  long long ne[N], h[N], e[N], idx = 1;
  long long dp[N][N], cnt[N][N];
  void add(long long a, long long b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
  void dfs(long long u, long long fa) {
    cnt[u][0] = 1;
    for (long long i = h[u]; i; i = ne[i]) {
      long long j = e[i];
      if (j == fa) continue;
      dfs(j, u);
      for (long long deep = 1; deep <= n; deep++)
        cnt[u][deep] += cnt[j][deep - 1];
    }
  }
  long long slove(long long u) {
    for (long long i = 0; i <= n; i++)
      for (long long j = 0; j <= n; j++) cnt[i][j] = 0;
    dfs(u, -1);
    vector<long long> son;
    son.push_back(-1);
    for (long long i = h[u]; i; i = ne[i]) son.push_back(e[i]);
    long long len = (long long)son.size(), res = 0;
    for (long long d = 0; d <= n; d++) {
      dp[0][0] = 1;
      for (long long i = 1; i < len; i++) {
        dp[i][0] = 1;
        for (long long j = 1; j <= i; j++) {
          dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
          dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * cnt[son[i]][d] % mod)) % mod;
        }
      }
      res = (res + dp[len - 1][k]) % mod;
      for (long long i = 1; i < len; i++)
        for (long long j = 1; j <= i; j++) dp[i][j] = 0;
    }
    return res;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tt;
    cin >> tt;
    while (tt--) {
      for (long long i = 1; i <= idx; i++) h[i] = 0;
      idx = 1;
      cin >> n >> k;
      for (long long i = 1; i < n; i++) {
        long long a, b;
        cin >> a >> b;
        add(a, b);
        add(b, a);
      }
      long long ans = 0;
      if (k == 2)
        ans = (n * (n - 1) / 2) % mod;
      else
        for (long long root = 1; root <= n; root++)
          ans = (ans + slove(root)) % mod;
      cout << ans << endl;
    }
    return 0;
  }
  