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;
  template <typename T1, typename T2>
  istream &operator>>(istream &in, pair<T1, T2> &a) {
    in >> a.first >> a.second;
    return in;
  }
  template <typename T1, typename T2>
  ostream &operator<<(ostream &out, pair<T1, T2> a) {
    out << a.first << " " << a.second;
    return out;
  }
  template <typename T, typename T1>
  T maxs(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T mins(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  const long long MOD = 1000000007;
  struct mod_int {
    long long val;
    mod_int(long long v = 0) {
      if (v < 0) v = v % MOD + MOD;
      if (v >= MOD) v %= MOD;
      val = v;
    }
    static long long mod_inv(long long a, long long m = MOD) {
      long long g = m, r = a, first = 0, second = 1;
      while (r != 0) {
        long long q = g / r;
        g %= r;
        swap(g, r);
        first -= q * second;
        swap(first, second);
      }
      return first < 0 ? first + m : first;
    }
    explicit operator long long() const { return val; }
    mod_int &operator+=(const mod_int &other) {
      val += other.val;
      if (val >= MOD) val -= MOD;
      return *this;
    }
    mod_int &operator-=(const mod_int &other) {
      val -= other.val;
      if (val < 0) val += MOD;
      return *this;
    }
    static unsigned fast_mod(uint64_t first, unsigned m = MOD) {
      return first % m;
      unsigned x_high = first >> 32, x_low = (unsigned)first;
      unsigned quot, rem;
      asm("divl %4\n" : "=a"(quot), "=d"(rem) : "d"(x_high), "a"(x_low), "r"(m));
      return rem;
    }
    mod_int &operator*=(const mod_int &other) {
      val = fast_mod((uint64_t)val * other.val);
      return *this;
    }
    mod_int &operator/=(const mod_int &other) { return *this *= other.inv(); }
    friend mod_int operator+(const mod_int &a, const mod_int &b) {
      return mod_int(a) += b;
    }
    friend mod_int operator-(const mod_int &a, const mod_int &b) {
      return mod_int(a) -= b;
    }
    friend mod_int operator*(const mod_int &a, const mod_int &b) {
      return mod_int(a) *= b;
    }
    friend mod_int operator/(const mod_int &a, const mod_int &b) {
      return mod_int(a) /= b;
    }
    mod_int &operator++() {
      val = val == MOD - 1 ? 0 : val + 1;
      return *this;
    }
    mod_int &operator--() {
      val = val == 0 ? MOD - 1 : val - 1;
      return *this;
    }
    mod_int operator++(int32_t) {
      mod_int before = *this;
      ++*this;
      return before;
    }
    mod_int operator--(int32_t) {
      mod_int before = *this;
      --*this;
      return before;
    }
    mod_int operator-() const { return val == 0 ? 0 : MOD - val; }
    bool operator==(const mod_int &other) const { return val == other.val; }
    bool operator!=(const mod_int &other) const { return val != other.val; }
    mod_int inv() const { return mod_inv(val); }
    mod_int pow(long long p) const {
      assert(p >= 0);
      mod_int a = *this, result = 1;
      while (p > 0) {
        if (p & 1) result *= a;
        a *= a;
        p >>= 1;
      }
      return result;
    }
    friend ostream &operator<<(ostream &stream, const mod_int &m) {
      return stream << m.val;
    }
    friend istream &operator>>(istream &stream, mod_int &m) {
      return stream >> m.val;
    }
  };
  long long solve() {
    long long n, k;
    cin >> n >> k;
    vector<vector<long long>> g(n + 1);
    for (long long i = 0; i + 1 < n; i++) {
      long long a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 2) {
      mod_int v = n;
      cout << v * (v - 1) / 2 << "\n";
      return 0;
    }
    function<void(long long, long long, long long, vector<long long> &)> dfs =
        [&](long long u, long long p, long long h, vector<long long> &d) {
          d[h]++;
          for (auto i : g[u]) {
            if (i != p) {
              dfs(i, u, h + 1, d);
            }
          }
        };
    mod_int ans = 0;
    auto solve = [&](long long r) {
      vector<vector<long long>> vals;
      for (auto i : g[r]) {
        vector<long long> d(n + 1);
        dfs(i, r, 1, d);
        vals.push_back(d);
      }
      for (long long d = 1; d <= n; d++) {
        vector<mod_int> dp(k + 1);
        dp[0] = 1;
        for (long long i = 0; i < vals.size(); i++) {
          vector<mod_int> new_dp(k + 1);
          for (long long j = 0; j <= k; j++) {
            new_dp[j] = dp[j];
            if (j >= 1) {
              new_dp[j] += dp[j - 1] * (vals[i][d]);
            }
          }
          swap(dp, new_dp);
        }
        ans += dp[k];
      }
    };
    for (long long i = 1; i <= n; i++) {
      solve(i);
    }
    cout << ans << "\n";
    return 0;
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  