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 mod = 1e9 + 7;
  long long add(long long a, long long b) { return (a + b) % mod; }
  long long sub(long long a, long long b) { return ((a - b) % mod + mod) % mod; }
  long long mul(long long a, long long b) { return (a * b) % mod; }
  long long power(long long a, long long b) {
    if (b == 0) return 1LL;
    long long tmp = power(a, b / 2);
    tmp = mul(tmp, tmp);
    if (b & 1) tmp = mul(tmp, a);
    return tmp;
  }
  const int maxn = 105;
  long long dp[maxn][maxn], fact[maxn], ifact[maxn], ans;
  vector<int> adjlist[maxn];
  int n, k, depth[maxn], groupsize[maxn][maxn];
  void precalc() {
    fact[0] = ifact[0] = 1;
    for (int i = 1; i < maxn; i++) {
      fact[i] = mul(i, fact[i - 1]);
      ifact[i] = power(fact[i], mod - 2);
    }
  }
  long long ncr(long long n, long long r) {
    if (r > n) return 0LL;
    return mul(fact[n], mul(ifact[n - r], ifact[r]));
  }
  void pivot(int p) {
    if (adjlist[p].size() < k) return;
    memset(depth, 0, sizeof(depth));
    memset(dp, 0, sizeof(dp));
    memset(groupsize, 0, sizeof(groupsize));
    depth[p] = 1;
    queue<pair<int, int>> q;
    for (auto u : adjlist[p]) {
      q.push({u, u});
      depth[u] = 2;
    }
    while (!q.empty()) {
      pair<int, int> cur = q.front();
      q.pop();
      groupsize[cur.second][depth[cur.first]]++;
      for (auto u : adjlist[cur.first]) {
        if (depth[u] != 0) continue;
        depth[u] = depth[cur.first] + 1;
        q.push({u, cur.second});
      }
    }
    for (int i = 0; i <= adjlist[p].size(); i++) {
      dp[i][0] = 1;
    }
    for (int d = 2; d < maxn; d++) {
      int tot = 0;
      for (auto u : adjlist[p]) {
        tot += min(groupsize[u][d], 1);
      }
      if (tot < k) break;
      for (int i = 0; i < adjlist[p].size(); i++) {
        for (int j = 1; j <= min(i + 1, k); j++) {
          dp[i + 1][j] =
              add(dp[i][j], mul(dp[i][j - 1], groupsize[adjlist[p][i]][d]));
        }
      }
      ans = add(ans, dp[adjlist[p].size()][k]);
    }
  }
  void solve() {
    ans = 0;
    memset(dp, 0, sizeof(dp));
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      adjlist[i].clear();
    }
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      adjlist[u].push_back(v);
      adjlist[v].push_back(u);
    }
    if (k == 2) {
      cout << ncr(n, 2);
      return;
    }
    for (int i = 1; i <= n; i++) {
      pivot(i);
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    precalc();
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  