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;
  long long t, n, ver[210], nxt[210], hed[210], tot = 0, k, f[110][110], dep[110],
                                                dp[110][110];
  void add(long long x, long long y) {
    ver[++tot] = y, nxt[tot] = hed[x], hed[x] = tot;
  }
  void pre() {
    tot = 0;
    memset(ver, 0, sizeof(ver));
    memset(nxt, 0, sizeof(nxt));
    memset(hed, 0, sizeof(hed));
  }
  void dfs(long long x) {
    for (long long i = hed[x]; i; i = nxt[i])
      if (!dep[ver[i]]) dep[ver[i]] = dep[x] + 1, dfs(ver[i]);
  }
  void dbfs(long long x, long long num) {
    f[num][dep[x]]++;
    for (long long i = hed[x]; i; i = nxt[i])
      if (dep[ver[i]] > dep[x]) dbfs(ver[i], num);
  }
  long long work(long long x) {
    memset(dp, 0, sizeof(dp));
    memset(f, 0, sizeof(f));
    memset(dep, 0, sizeof(dep));
    long long cnt = 0, anss = 0;
    dep[x] = 1;
    dfs(x);
    for (int i = hed[x]; i; i = nxt[i]) dbfs(ver[i], ++cnt);
    for (int i = 2; i <= n; ++i) {
      long long cntt = 0;
      for (int j = 1; j <= cnt; ++j)
        if (f[j][i]) cntt++;
      if (cntt < k) break;
      for (int j = 0; j <= cnt; ++j) dp[j][0] = 1;
      for (long long j = 1; j <= cnt; ++j)
        for (long long l = 1; l <= j; ++l)
          dp[j][l] = (dp[j - 1][l] + dp[j - 1][l - 1] * f[j][i] % 1000000007) %
                     1000000007;
      anss = (anss + dp[cnt][k]) % 1000000007;
    }
    return anss;
  }
  int main() {
    scanf("%lld", &t);
    while (t--) {
      pre();
      long long ans = 0;
      scanf("%lld%lld", &n, &k);
      for (long long i = 1, x, y; i < n; ++i)
        scanf("%lld%lld", &x, &y), add(x, y), add(y, x);
      if (k == 2)
        printf("%lld\n", n * (n - 1) / 2);
      else {
        for (long long i = 1; i <= n; ++i) ans = (ans + work(i)) % 1000000007;
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  