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;
  using P = pair<ll, ll>;
  using PP = pair<ll, P>;
  const ll n_ = 111 + 100, inf = 1e18, mod = 1e9 + 7, sqrtN = 333, p = 27;
  ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
  ll n, m, k, tc = 1, a, b, c, d, sum, x, y, z, w, base, ans;
  ll gcd(ll x, ll y) {
    if (!y) return x;
    return gcd(y, x % y);
  }
  vector<ll> v[n_];
  ll checked[n_], cnt[n_], dp[n_][n_];
  ll f(ll x, ll y) {
    for (int i = 0; i <= x; i++)
      for (int j = 0; j <= y; j++) dp[i][j] = 0;
    dp[0][0] = 1;
    for (int i = 0; i < x; i++)
      for (int j = 0; j <= y; j++) {
        dp[i + 1][j] = (dp[i + 1][j] + dp[i][j]) % mod;
        dp[i + 1][j + 1] = (dp[i + 1][j + 1] + (dp[i][j] * cnt[i]) % mod) % mod;
      }
    return dp[x][y];
  }
  void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) v[i].clear();
    for (int i = 0; i < n - 1; i++) {
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    ans = 0;
    if (k == 2) {
      ans = n * (n - 1) / 2;
      cout << ans % mod << '\n';
      return;
    }
    for (int i = 1; i <= n; i++) {
      memset(checked, 0, sizeof(checked));
      checked[i] = true;
      vector<P> temp;
      for (int j = 0; j < v[i].size(); j++) {
        temp.push_back({v[i][j], j});
        cnt[j] = 1;
        checked[v[i][j]] = true;
      }
      while (temp.size()) {
        ans += f((ll)v[i].size(), k);
        ans %= mod;
        vector<P> V;
        for (auto nxt : temp) {
          cnt[nxt.second]--;
          for (auto nxt2 : v[nxt.first]) {
            if (checked[nxt2]) continue;
            checked[nxt2] = true;
            cnt[nxt.second]++;
            V.push_back({nxt2, nxt.second});
          }
        }
        temp = V;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    cin >> tc;
    while (tc--) solve();
  }
  