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 = 100 + 7;
  const long long MOD = 1E9 + 7;
  long long res = 0, fact[N], rev[N], dp[N][N];
  vector<int> G[N];
  int timer = 0, cnt[N], tin[N], depth[N], sum[N];
  void add(int pos, int val) {
    if (tin[pos] != timer) {
      tin[pos] = timer;
      cnt[pos] = 0;
    }
    cnt[pos] += val;
  }
  int get(int pos) {
    if (tin[pos] != timer) {
      tin[pos] = timer;
      cnt[pos] = 0;
    }
    return cnt[pos];
  }
  void dfs(int v, int par) {
    depth[v] = depth[par] + 1;
    add(depth[v], 1);
    for (int to : G[v]) {
      if (to == par) continue;
      dfs(to, v);
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      G[u].push_back(v);
      G[v].push_back(u);
    }
    res = 0;
    for (int root = 1; root <= n; ++root) {
      if (G[root].size() < k) continue;
      for (int i = 0; i <= n; ++i) {
        for (int j = 1; j <= k; ++j) dp[i][j] = 0;
        dp[i][0] = 1;
        sum[i] = 0;
      }
      depth[root] = 0;
      for (int v : G[root]) {
        ++timer;
        dfs(v, root);
        for (int dep = 1; dep <= n; ++dep) {
          if (get(dep) == 0) continue;
          for (int ct = min(k - 1, sum[dep]); ct >= 0; --ct) {
            dp[dep][ct + 1] = (dp[dep][ct + 1] + dp[dep][ct] * get(dep)) % MOD;
          }
          sum[dep] = min(sum[dep] + 1, k);
        }
      }
      for (int dep = 1; dep <= n; ++dep) res = (res + dp[dep][k]) % MOD;
    }
    if (k == 2) {
      res = long long(n) * (n - 1) / 2 % MOD;
    }
    if (k == 1) {
      res = n;
    }
    cout << res << endl;
    for (int i = 1; i <= n; ++i) G[i].clear();
  }
  long long binpow(long long x, int power) {
    long long mult = x % MOD;
    x = 1;
    while (power) {
      if (power & 1) x = x * mult % MOD;
      mult = mult * mult % MOD;
      power = power >> 1;
    }
    return x;
  }
  int main() {
    fact[0] = rev[0] = 1;
    for (int i = 1; i < N; ++i) {
      fact[i] = fact[i - 1] * i % MOD;
      rev[i] = binpow(fact[i], MOD - 2);
    }
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  