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;
  int T, mod = 1e9 + 7, n, k, deep[N], timer, sz;
  long long ans, fact[N], mem[N][N], vis[N][N];
  vector<int> adj[N], v;
  map<int, int> mp[N];
  void _clear() {
    ans = 0;
    for (int i = 1; i <= n; i++) adj[i].clear();
  }
  long long po(long long x, long long os) {
    if (os == 0) return 1;
    long long z = po(x, os / 2);
    if (os & 1) return z * z % mod * x % mod;
    return z * z % mod;
  }
  long long modInv(long long x) { return po(x, mod - 2); }
  long long c(long long n, long long r) {
    return fact[n] * modInv(fact[r] * fact[n - r] % mod) % mod;
  }
  void dfs(int u, int v) {
    deep[u] = deep[v] + 1;
    mp[u][deep[u]]++;
    for (auto &x : adj[u])
      if (x != v) {
        dfs(x, u);
        for (auto &p : mp[x]) mp[u][p.first] += p.second;
      }
  }
  long long dp(int i, int k) {
    if (k == 0) return 1;
    if (i == sz) return 0;
    long long &ret = mem[i][k];
    if (vis[i][k] == timer) return ret;
    vis[i][k] = timer;
    return ret = (dp(i + 1, k) + dp(i + 1, k - 1) * v[i] % mod) % mod;
  }
  long long calc(int p) {
    deep[p] = 0;
    for (int i = 1; i <= n; i++) mp[i].clear();
    dfs(p, p);
    long long ret = 0;
    for (int i = 2; i <= n; i++) {
      int num = 0;
      v.clear();
      for (auto &x : adj[p])
        if (mp[x].find(i) != mp[x].end()) {
          num++;
          v.push_back(mp[x][i]);
        }
      if (num < k) continue;
      timer++;
      sz = (int)(v.size());
      ret = (ret + dp(0, k)) % mod;
    }
    return ret;
  }
  int main() {
    fact[0] = 1;
    for (int i = 1; i < N; i++) fact[i] = fact[i - 1] * i % mod;
    scanf("%d", &T);
    while (T--) {
      scanf("%d%d", &n, &k);
      _clear();
      for (int i = 1; i <= n - 1; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        adj[a].push_back(b);
        adj[b].push_back(a);
      }
      if (k == 2) {
        printf("%lld\n", c(n, 2));
        continue;
      }
      for (int i = 1; i <= n; i++) ans = (ans + calc(i)) % mod;
      printf("%lld\n", ans);
    }
  }
  