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 long long p = 1e9 + 7;
  const long long maxn = 105;
  vector<long long> a[maxn];
  bool used[maxn];
  long long l[maxn][maxn];
  long long ans;
  long long c;
  long long k;
  long long slv1(vector<long long> v, long long k) {
    long long dp[v.size() + 1][k + 1];
    for (long long i = 0; i <= v.size(); ++i)
      for (long long j = 0; j <= k; ++j) dp[i][j] = 0;
    dp[0][0] = 1;
    for (long long i = 0; i < v.size(); ++i)
      for (long long j = 0; j <= k; ++j) {
        if (j != k) {
          dp[i + 1][j + 1] += (dp[i][j] * v[i]);
          dp[i + 1][j + 1] %= p;
        }
        {
          dp[i + 1][j] += (dp[i][j]);
          dp[i + 1][j] %= p;
        }
      }
    return dp[v.size()][k];
  }
  void dfs(long long x) {
    for (long long i = 0; i < maxn; ++i) l[x][i] = 0;
    l[x][0]++;
    used[x] = true;
    vector<long long> z;
    for (auto v : a[x]) {
      if (used[v]) continue;
      dfs(v);
      z.push_back(v);
      for (long long i = 0; i < maxn; ++i) l[x][i + 1] += l[v][i];
    }
    used[x] = false;
    if (x == c) {
      for (long long i = 0; i < maxn; ++i) {
        vector<long long> v;
        for (auto h : z) {
          v.push_back(l[h][i]);
        }
        long long o = slv1(v, k);
        ans += o;
        ans %= p;
        ans += p;
        ans %= p;
      }
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n >> k;
      ans = 0;
      for (long long i = 0; i < maxn; ++i) {
        a[i].clear();
        used[i] = false;
      }
      for (long long i = 0; i < (n - 1); ++i) {
        long long x, y;
        cin >> x >> y;
        x--;
        y--;
        a[x].push_back(y);
        a[y].push_back(x);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
        continue;
      }
      for (long long i = 0; i < n; ++i) {
        c = i;
        dfs(i);
      }
      cout << ans << endl;
    }
    return 0;
  }
  