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 int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
  const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
            dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
  const long long mod = 1000000007;
  const int base = 311;
  const int N = 105;
  int n, k;
  vector<int> adj[N];
  int h[N];
  void dfs(int u, int pre, int &cnt, int len) {
    if (h[u] == len) cnt++;
    for (int v : adj[u])
      if (v != pre) {
        h[v] = h[u] + 1;
        dfs(v, u, cnt, len);
      }
  }
  void add(int &a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  void gogo() {
    cin >> n >> k;
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    int ans = 0;
    if (k == 2)
      cout << 1LL * n * (n - 1) / 2 << '\n';
    else {
      for (int len = 0; len <= n; ++len) {
        for (int i = 1; i <= n; ++i) {
          memset(h, 0, sizeof(h));
          vector<int> all;
          for (int u : adj[i]) {
            int cnt = 0;
            dfs(u, i, cnt, len);
            all.push_back(cnt);
          }
          vector<int> dp(k + 5);
          dp[0] = 1;
          for (int ptr = 0; ptr < (int)all.size(); ++ptr) {
            vector<int> f(k + 5, 0);
            for (int j = 0; j <= k; ++j) {
              if (dp[j] == 0) continue;
              add(f[j], dp[j]);
              add(f[j + 1], (long long)dp[j] * all[ptr] % mod);
            }
            swap(dp, f);
          }
          add(ans, dp[k]);
        }
      }
      cout << ans << '\n';
    }
    for (int i = 1; i <= n; ++i) adj[i].clear();
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    if (fopen("sol"
              ".inp",
              "r")) {
      freopen(
          "sol"
          ".inp",
          "r", stdin);
      freopen(
          "sol"
          ".out",
          "w", stdout);
    }
    int t;
    cin >> t;
    while (t--) gogo();
  }
  