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;
  using ll = long long int;
  int t;
  int n, k;
  vector<int> v[100 + 1];
  ll dp[100 + 1][100 + 1];
  bool vst[100 + 1];
  ll ans;
  void dfs(int now, int cnt, int root) {
    vst[now] = true;
    dp[root][now] = cnt;
    for (auto there : v[now]) {
      if (!vst[there]) {
        dfs(there, cnt + 1, root);
      }
    }
  }
  ll c[128][128];
  ll add(ll x, ll y) { return (x + y) % (ll)1000000007; }
  ll mul(ll x, ll y) { return x * y % (ll)1000000007; }
  ll rundp(vector<int>& arr, int m, int k) {
    memset(c, 0, sizeof c);
    c[0][0] = 1;
    for (int i = 0; i < m; i++) {
      for (int j = 0; j <= k; j++) {
        c[i + 1][j] = add(c[i + 1][j], c[i][j]);
        c[i + 1][j + 1] = add(c[i + 1][j + 1], mul(c[i][j], arr[i]));
      }
    }
    return c[m][k];
  }
  void solve(int now) {
    if (v[now].size() < k) return;
    memset(vst, false, sizeof vst);
    memset(dp, 0, sizeof dp);
    vst[now] = true;
    for (auto there : v[now]) {
      dfs(there, 1, there);
    }
    map<int, vector<int>> m;
    for (auto there : v[now]) {
      map<int, int> tmp;
      for (int i = 1; i <= n; i++) {
        if (dp[there][i] > 0) tmp[dp[there][i]]++;
      }
      for (auto it = tmp.begin(); it != tmp.end(); it++) {
        m[it->first].push_back(it->second);
      }
    }
    for (auto it = m.begin(); it != m.end(); it++) {
      if (it->second.size() >= k) {
        ans = (ans + rundp(it->second, it->second.size(), k)) % (ll)1000000007;
      }
    }
  }
  int comb[100 + 1][100 + 1];
  int pascal(int i, int j) {
    if (comb[i][j] != 0) {
      return comb[i][j];
    }
    if (j == 0 || j == i) {
      comb[i][j] = 1;
      return comb[i][j];
    }
    return comb[i][j] =
               (pascal(i - 1, j) + pascal(i - 1, j - 1)) % (ll)1000000007;
  }
  int main(void) {
    ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      memset(dp, 0, sizeof(dp));
      for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
      }
      ans = 0;
      for (int i = 1; i <= n; i++) {
        solve(i);
      }
      if (k == 2) {
        ans = pascal(n, k);
      }
      cout << ans << "\n";
      for (int i = 1; i <= n; i++) {
        v[i].clear();
        vector<int>().swap(v[i]);
      }
    }
    return 0;
  }
  