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;
  const ll A = 26;
  const ll inf = 1e18;
  const ll mod = 1e9 + 7;
  const ll BASE = 10000;
  double q;
  using ll = long long;
  inline ll bit(ll x, ll y) { return (x & (1ll << y)); }
  vector<ll> a[101];
  ll dp[101][101];
  ll sum[10], Dp[10][10];
  ll par[101];
  ll n, k, kq;
  void dfs(ll u) {
    dp[u][0] = 1;
    for (auto x : a[u])
      if (x != par[u]) {
        par[x] = u;
        dfs(x);
        for (ll i = 1; i <= n; i++)
          dp[u][i] = (dp[u][i] % mod + dp[x][i - 1] % mod) % mod;
      }
  }
  inline ll powmod(ll a, ll b) {
    ll res = 1ll;
    a %= mod;
    while (b) {
      if (b & 1ll) res = (res * a) % mod;
      a = (a * a) % mod;
      b >>= 1ll;
    }
    return res % mod;
  }
  ll d[100006], dk[100006];
  void init() {
    d[0] = d[1] = dk[0] = dk[1] = 1ll;
    for (ll i = 0; i <= 100000; i++) {
      dk[i] = powmod(i, mod - 2);
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    init();
    ll t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (ll i = 1; i <= n; i++) {
        a[i].clear();
      }
      for (ll i = 1; i <= n - 1; i++) {
        ll u, v;
        cin >> u >> v;
        a[u].push_back(v);
        a[v].push_back(u);
      }
      if (k == 2) {
        cout << ((n - 1) * n) / 2 << endl;
        continue;
      }
      kq = 0;
      for (ll i = 1; i <= n; i++)
        if (a[i].size() >= k) {
          for (ll j = 1; j <= n; j++) par[j] = j;
          for (ll j = 1; j <= n; j++)
            for (ll jj = 1; jj <= n; jj++) {
              dp[j][jj] = 0;
            }
          dfs(i);
          for (ll j = 0; j <= n - 1; j++) {
            vector<vector<ll>> Dp(n + 1, vector<ll>(n + 1));
            vector<ll> sum(k + 1, 0ll);
            for (auto x : a[i]) {
              Dp[x][1] = dp[x][j] % mod;
              sum[1] += dp[x][j] % mod;
              sum[1] %= mod;
            }
            for (ll jj = 2; jj <= k; jj++)
              for (auto jjj : a[i]) {
                Dp[jjj][jj] = (((sum[jj - 1] * dk[jj - 1]) % mod -
                                Dp[jjj][jj - 1] + mod + mod) %
                               mod * dp[jjj][j] % mod) %
                              mod;
                sum[jj] += Dp[jjj][jj];
                sum[jj] %= mod;
              }
            kq = (kq + (sum[k] * dk[k]) % mod) % mod;
          }
        }
      cout << kq << endl;
    }
  }
  