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;
  long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
  long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
  class pa3 {
   public:
    long long x;
    long long y, z;
    pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
    bool operator<(const pa3& p) const {
      if (x != p.x) return x < p.x;
      if (y != p.y) return y < p.y;
      return z < p.z;
    }
    bool operator>(const pa3& p) const {
      if (x != p.x) return x > p.x;
      if (y != p.y) return y > p.y;
      return z > p.z;
    }
    bool operator==(const pa3& p) const {
      return x == p.x && y == p.y && z == p.z;
    }
    bool operator!=(const pa3& p) const {
      return !(x == p.x && y == p.y && z == p.z);
    }
  };
  class pa4 {
   public:
    long long x;
    long long y, z, w;
    pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
        : x(x), y(y), z(z), w(w) {}
    bool operator<(const pa4& p) const {
      if (x != p.x) return x < p.x;
      if (y != p.y) return y < p.y;
      if (z != p.z) return z < p.z;
      return w < p.w;
    }
    bool operator>(const pa4& p) const {
      if (x != p.x) return x > p.x;
      if (y != p.y) return y > p.y;
      if (z != p.z) return z > p.z;
      return w > p.w;
    }
    bool operator==(const pa4& p) const {
      return x == p.x && y == p.y && z == p.z && w == p.w;
    }
  };
  class pa2 {
   public:
    long long x, y;
    pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
    pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
    pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
    bool operator<(const pa2& p) const { return y != p.y ? y < p.y : x < p.x; }
    bool operator>(const pa2& p) const { return x != p.x ? x < p.x : y < p.y; }
    bool operator==(const pa2& p) const {
      return abs(x - p.x) == 0 && abs(y - p.y) == 0;
    }
    bool operator!=(const pa2& p) const {
      return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
    }
  };
  string itos(long long i) {
    ostringstream s;
    s << i;
    return s.str();
  }
  long long Gcd(long long v, long long b) {
    if (v == 0) return b;
    if (b == 0) return v;
    if (v > b) return Gcd(b, v);
    if (v == b) return b;
    if (b % v == 0) return v;
    return Gcd(v, b % v);
  }
  long long extgcd(long long a, long long b, long long& x, long long& y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long d = extgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
  }
  pair<long long, long long> operator+(const pair<long long, long long>& l,
                                       const pair<long long, long long>& r) {
    return {l.first + r.first, l.second + r.second};
  }
  pair<long long, long long> operator-(const pair<long long, long long>& l,
                                       const pair<long long, long long>& r) {
    return {l.first - r.first, l.second - r.second};
  }
  ostream& operator<<(ostream& os, const vector<long long>& VEC) {
    for (auto v : VEC) os << v << " ";
    return os;
  }
  ostream& operator<<(ostream& os, const pair<long long, long long>& PAI) {
    os << PAI.first << " : " << PAI.second;
    return os;
  }
  ostream& operator<<(ostream& os, const pa3& PAI) {
    os << PAI.x << " : " << PAI.y << " : " << PAI.z;
    return os;
  }
  ostream& operator<<(ostream& os, const pa4& PAI) {
    os << PAI.x << " : " << PAI.y << " : " << PAI.z << " : " << PAI.w;
    return os;
  }
  ostream& operator<<(ostream& os,
                      const vector<pair<long long, long long> >& VEC) {
    for (auto v : VEC) os << v << " ";
    return os;
  }
  ostream& operator<<(ostream& os, const vector<pa3>& VEC) {
    for (auto v : VEC) {
      os << v << " ";
      os << endl;
    }
    return os;
  }
  long long beki(long long wa, long long rr, long long warukazu) {
    if (rr == 0) return 1 % warukazu;
    if (rr == 1) return wa % warukazu;
    wa %= warukazu;
    if (rr % 2 == 1)
      return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
    long long zx = beki(wa, rr / 2, warukazu);
    return (zx * zx) % warukazu;
  }
  long long pr[2521000];
  long long inv[2521000];
  const long long mod = 1000000007;
  long long comb(long long nn, long long rr) {
    if (rr < 0 || rr > nn || nn < 0) return 0;
    long long r = pr[nn] * inv[rr];
    r %= mod;
    r *= inv[nn - rr];
    r %= mod;
    return r;
  }
  void gya(long long ert) {
    pr[0] = 1;
    for (long long i = 1; i <= ert; i++) {
      pr[i] = ((long long)pr[i - 1] * i) % mod;
    }
    inv[ert] = beki(pr[ert], mod - 2, mod);
    for (long long i = ert - 1; i >= 0; i--) {
      inv[i] = (long long)inv[i + 1] * (i + 1) % mod;
    }
  }
  long long beki2(long long wa, long long rr) {
    if (rr == 0) return 1;
    if (rr == 1) return wa;
    if (rr % 2 == 1) return ((long long)beki2(wa, rr - 1) * (long long)wa) % mod;
    long long zx = beki2(wa, rr / 2);
    return (zx * zx) % mod;
  }
  long long beki3(long long a, long long b) {
    long long ANS = 1;
    long long be = a;
    while (b) {
      if (b & 1) {
        ANS *= be;
        ANS %= mod;
      }
      be *= be;
      be %= mod;
      b /= 2;
    }
    return ANS;
  }
  vector<long long> G[20002];
  long long d[200][200] = {};
  void dfs(long long r, long long p, long long di, long long it) {
    d[it][di]++;
    for (auto v : G[r])
      if (v != p) dfs(v, r, di + 1, it);
  }
  long long dp[200][200] = {};
  long long solve(vector<long long> ve, long long k) {
    long long n = ve.size();
    for (long long i = 0; i <= n; i++)
      for (long long j = 0; j <= n; j++) dp[i][j] = 0;
    dp[0][0] = 1;
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j <= i; j++) {
        dp[i + 1][j] += dp[i][j];
        dp[i + 1][j] %= mod;
        dp[i + 1][j + 1] += dp[i][j] * ve[i] % mod;
        dp[i + 1][j + 1] %= mod;
      }
    return dp[n][k];
  }
  long long solve(long long r, long long k) {
    if ((long long)G[r].size() < k) return 0;
    long long it = 0;
    for (auto v : G[r]) {
      dfs(v, r, 0, it);
      it++;
    }
    long long e = G[r].size();
    long long ans = 0;
    for (long long kk = 0; kk < 110; kk++) {
      long long ko = 0;
      vector<long long> ve;
      for (long long j = 0; j < e; j++) {
        if (d[j][kk]) ko++, ve.push_back(d[j][kk]);
      }
      if (ko < k) break;
      ans += solve(ve, k);
    }
    for (long long i = 0; i <= 110; i++)
      for (long long j = 0; j < e; j++) d[j][i] = 0;
    return ans % mod;
  }
  long long ke(long long n) {
    for (long long i = 0; i < n; i++) G[i].clear();
    return 0;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n - 1; i++) {
      long long y, yy;
      cin >> y >> yy;
      y--, yy--;
      G[y].push_back(yy);
      G[yy].push_back(y);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 % mod << endl;
      ke(n);
      return;
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++) ans += solve(i, k);
    ke(n);
    cout << ans % mod << endl;
  }
  signed main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    long long n = 1;
    cin >> n;
    for (long long i = 0; i < n; i++) solve();
    return 0;
  }
  