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 N = 110;
  const long long mod = 1e9 + 7LL;
  vector<int> lista[N];
  int vis[N], g[N], peso[N], qt;
  long long dp[N][N], visdp[N][N], t = 0;
  long long solve(int pos, int k) {
    if (!k) return 1;
    if (pos == qt) return 0;
    long long &ans = dp[pos][k];
    if (visdp[pos][k] == t) return ans;
    visdp[pos][k] = t;
    ans = (peso[pos] * solve(pos + 1, k - 1) + solve(pos + 1, k)) % mod;
    return ans;
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, c;
    int q;
    cin >> q;
    while (q--) {
      cin >> n >> c;
      for (int i = 0; i <= n; ++i) {
        vis[i] = 0;
        lista[i].clear();
      }
      for (int i = 0; i < n - 1; ++i) {
        int a, b;
        cin >> a >> b;
        lista[a].push_back(b);
        lista[b].push_back(a);
      }
      long long ans = 0;
      if (c == 2) {
        cout << n * (n - 1) / 2 << "\n";
        continue;
      }
      for (int i = 1; i <= n; ++i) {
        queue<int> f, f1;
        vis[i] = i;
        qt = 0;
        for (int u : lista[i]) {
          g[u] = qt++;
          peso[g[u]] = 1;
          vis[u] = i;
          f.push(u);
        }
        t++;
        ans = (ans + solve(0, c)) % mod;
        while (!f.empty()) {
          while (!f.empty()) {
            int u = f.front();
            f.pop();
            peso[g[u]]--;
            for (int v : lista[u])
              if (vis[v] != i) {
                g[v] = g[u];
                peso[g[v]]++;
                vis[v] = i;
                f1.push(v);
              }
          }
          t++;
          ans = (ans + solve(0, c)) % mod;
          swap(f1, f);
        }
      }
      cout << ans << "\n";
    }
    return 0;
  }
  