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;
  using ld = long double;
  template <typename T1, typename T2>
  inline void chkmin(T1& x, const T2& y) {
    if (y < x) x = y;
  }
  template <typename T1, typename T2>
  inline void chkmax(T1& x, const T2& y) {
    if (x < y) x = y;
  }
  mt19937 rnd(time(0));
  const int MAXN = 10e3 + 228;
  vector<int> g[MAXN];
  int d[MAXN];
  int mod = 1000000007;
  int mul(int a, int b) { return 1ll * a * b % mod; }
  int add(int a, int b) {
    if (a + b >= mod) return a + b - mod;
    return a + b;
  }
  int sub(int a, int b) {
    if (a - b < 0) return a - b + mod;
    return a - b;
  }
  int binpow(int a, int b) {
    int res = 1;
    while (b) {
      if (b & 1) res = mul(res, a);
      a = mul(a, a);
      b >>= 1;
    }
    return res;
  }
  int inv(int x) { return binpow(x, mod - 2); }
  int fact[MAXN];
  int rfact[MAXN];
  int cnk(int n, int k) {
    if (n < 0 || k > n) return 0;
    return mul(fact[n], mul(rfact[k], rfact[n - k]));
  }
  int tpc[MAXN];
  void dfs(int v, int p, int tp) {
    d[v] = d[p] + 1;
    tpc[v] = tp;
    for (int i : g[v]) {
      if (i != p) {
        int ntp = tp;
        if (v == p) ntp = i;
        dfs(i, v, ntp);
      }
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cout.precision(20), cout.setf(ios::fixed);
    fact[0] = 1;
    for (int i = 1; i < MAXN; ++i) fact[i] = mul(fact[i - 1], i);
    rfact[MAXN - 1] = inv(fact[MAXN - 1]);
    for (int i = MAXN - 2; i >= 0; --i) rfact[i] = mul(rfact[i + 1], i + 1);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; ++i) g[i].clear();
      for (int i = 0; i < n - 1; ++i) {
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        continue;
      }
      int ans = 0;
      for (int i = 1; i <= n; ++i) {
        fill(d, d + n + 1, 0);
        dfs(i, i, -1);
        for (int dd = 2; dd <= n; ++dd) {
          vector<int> cnts(n + 1);
          for (int j = 1; j <= n; ++j) {
            if (d[j] == dd) {
              cnts[tpc[j]]++;
            }
          }
          vector<int> itms;
          for (int i : cnts) {
            if (i) itms.push_back(i);
          }
          vector<int> dp(n + 1, 0);
          dp[0] = 1;
          for (int j : itms) {
            vector<int> ndp = dp;
            for (int x = 1; x <= 1; ++x) {
              for (int gg = 0; gg + x <= n; ++gg) {
                ndp[gg + x] = add(ndp[gg + x], mul(dp[gg], cnk(j, x)));
              }
            }
            swap(dp, ndp);
          }
          ans = add(ans, dp[k]);
        }
      }
      cout << ans << '\n';
    }
  }
  