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;
  void io() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
  }
  const int32_t MOD = 1e9 + 7;
  int32_t add(int32_t a, int32_t b) {
    if (a + b < MOD) return a + b;
    return a + b - MOD;
  }
  int32_t sub(int32_t a, int32_t b) {
    if (a - b >= 0) return a - b;
    return a - b + MOD;
  }
  int32_t mul(int32_t a, int32_t b) { return int64_t(a) * b % MOD; }
  vector<vector<int32_t>> g;
  int32_t n;
  void dfs(vector<vector<int32_t>> &cntH, int32_t v, int32_t p) {
    cntH[v][0] = 1;
    for (auto &u : g[v]) {
      if (u != p) {
        dfs(cntH, u, v);
        for (int32_t i = 1; i < n; ++i) {
          cntH[v][i] += cntH[u][i - 1];
        }
      }
    }
  }
  void solve() {
    int32_t k;
    cin >> n >> k;
    int32_t ans = 0;
    g.clear();
    g.resize(n);
    for (int32_t i = 0; i + 1 < n; ++i) {
      int32_t a, b;
      cin >> a >> b;
      --a, --b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      return;
    }
    for (int32_t v = 0; v < n; ++v) {
      vector<vector<int32_t>> cntH(n, vector<int32_t>(n, 0));
      dfs(cntH, v, v);
      int32_t m = g[v].size();
      for (int32_t h = 0; h < n; ++h) {
        vector<int32_t> cnt;
        cnt.reserve(m);
        for (auto &u : g[v]) {
          cnt.push_back(cntH[u][h]);
        }
        vector<vector<int32_t>> dp(m + 1, vector<int32_t>(k + 1, 0));
        dp[0][0] = 1;
        for (int32_t i = 0; i < m; ++i) {
          dp[i + 1][0] = 1;
          for (int32_t j = 1; j <= k; ++j) {
            dp[i + 1][j] =
                add(dp[i + 1][j], add(dp[i][j], mul(dp[i][j - 1], cnt[i])));
          }
        }
        ans = add(ans, dp[m][k]);
      }
    }
    cout << ans << '\n';
  }
  int32_t main() {
    io();
    int32_t ttt;
    std::cin >> ttt;
    while (ttt-- > 0) {
      solve();
    }
    return 0;
  }
  