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 = 105;
  int n, d[N], k, vis[N][N], dp[N][N], id, mod = 1e9 + 7, cnt[N][N];
  vector<int> adj[N], v;
  void dfs(int node, int par, int parCh) {
    if (parCh) {
      cnt[parCh][d[node]]++;
    }
    for (auto ch : adj[node]) {
      if (ch == par) continue;
      d[ch] = d[node] + 1;
      if (node == par)
        dfs(ch, node, ch);
      else
        dfs(ch, node, parCh);
    }
  }
  int solve(int i, int rem) {
    if (!rem) return 1;
    if (i == (int)v.size()) return 0;
    if (vis[i][rem] == id) return dp[i][rem];
    vis[i][rem] = id;
    int ans = solve(i + 1, rem);
    if (v[i]) ans = (ans + 1ll * v[i] * solve(i + 1, rem - 1)) % mod;
    return dp[i][rem] = ans;
  }
  void clear() {
    for (int i = 0; i <= n; i++) {
      adj[i].clear();
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (int i = 1; i < n; i++) {
        int x, y;
        cin >> x >> y;
        adj[x].push_back(y);
        adj[y].push_back(x);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        clear();
        continue;
      }
      long long ans = 0;
      for (int i = 1; i <= n; i++) {
        memset(d, 0, sizeof(d));
        memset(cnt, 0, sizeof(cnt));
        dfs(i, i, 0);
        for (int ds = 1; ds <= n; ds++) {
          v.clear();
          for (int ch = 1; ch <= n; ch++) {
            v.push_back(cnt[ch][ds]);
          }
          id++;
          ans = (ans + solve(0, k)) % mod;
        }
      }
      cout << ans << '\n';
      clear();
    }
    return 0;
  }
  