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 ull = unsigned long long int;
  const int maxx = 105;
  const int mod = 1000000007;
  ll power(ll n, ll first, ll mod) {
    if (first < 0) return 0ll;
    if (first == 0) return 1ll;
    ll cur = power(n, first / 2, mod);
    if (first & 1) return (cur % mod * cur % mod * n) % mod;
    return (cur % mod * cur % mod) % mod;
  }
  ll mul(ll a, ll b, ll mod) {
    if (b == 0) return 0ll;
    ll cur = mul(a, b / 2, mod);
    if (b & 1) {
      return (2 * cur + a) % mod;
    }
    return (2 * cur) % mod;
  }
  ll gcd(ll a, ll b, ll& first, ll& second) {
    if (b == 0) {
      first = 1;
      second = 0;
      return a;
    }
    ll x1, y1;
    ll g = gcd(b, a % b, x1, y1);
    first = y1;
    y1 = x1 - a / b * y1;
    return g;
  }
  vector<int> edge[maxx];
  ll f[maxx][maxx], h[maxx], visit[maxx], cnt[maxx];
  void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i <= n; ++i) {
      edge[i].clear();
    }
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    if (k == 2) {
      cout << (n * (n - 1) / 2) % mod << '\n';
      return;
    }
    ll ans = 0;
    for (int i = 1; i <= n; ++i) {
      memset(visit, 0, sizeof visit);
      visit[i] = 1;
      vector<pair<int, int> > tmp;
      int m = edge[i].size();
      for (int e = 0; e < m; ++e) {
        tmp.push_back({edge[i][e], e});
        cnt[e] = 1;
        visit[edge[i][e]] = 1;
      }
      while (1) {
        if (tmp.empty()) break;
        memset(f, 0, sizeof f);
        for (int j = 0; j <= m; ++j) f[j][0] = 1;
        for (int j = 1; j <= m; ++j) {
          for (int t = 1; t <= k; ++t) {
            f[j][t] = (f[j][t] + f[j - 1][t]) % mod;
            f[j][t] = (f[j][t] + cnt[j - 1] * f[j - 1][t - 1] % mod) % mod;
          }
        }
        ans = (ans + f[m][k]) % mod;
        vector<pair<int, int> > cur;
        for (auto p : tmp) {
          cnt[p.second]--;
          for (auto second : edge[p.first])
            if (!visit[second]) {
              cur.emplace_back(second, p.second);
              visit[second] = true;
              cnt[p.second]++;
            }
        }
        tmp = cur;
      }
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  