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
#include <bits/stdc++.h>
using namespace std;
void PV(vector<long long> v) {
  for (long long i = 0; i < (long long)v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void PVV(vector<pair<long long, long long> > v) {
  for (long long i = 0; i < (long long)v.size(); i++)
    cout << v[i].first << " " << v[i].second << "\n";
}
void PA(long long v[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cout << v[i] << ' ';
  cout << "\n";
}
void IN(long long a[], long long n, long long x = 0) {
  for (long long i = x; i < n + x; i++) cin >> a[i];
}
inline void op() {}
const long long M = 1e9 + 7;
const long long IM = 1e18 + 37;
const long long N = 1000 + 10;
long long factI[N + 1];
long long Numi[N + 1];
long long fact[N + 1];
void InverseofNumber(long long p = M) {
  Numi[0] = Numi[1] = 1;
  for (long long i = 2; i <= N; i++) Numi[i] = Numi[p % i] * (p - p / i) % p;
}
void InverseofFactorial(long long p = M) {
  factI[0] = factI[1] = 1;
  for (long long i = 2; i <= N; i++) factI[i] = (Numi[i] * factI[i - 1]) % p;
}
void factorial(long long p = M) {
  fact[0] = 1;
  for (long long i = 1; i <= N; i++) {
    fact[i] = (fact[i - 1] * i) % p;
  }
}
long long nPr(long long N, long long R, long long p = M) {
  if (N - R < 0 || R < 0) return 0;
  long long ans = ((fact[N]) % p * factI[N - R]) % p;
  return ans;
}
long long nCr(long long N, long long R, long long p = M) {
  if (N - R < 0 || R < 0) return 0;
  long long ans = ((fact[N] * factI[R]) % p * factI[N - R]) % p;
  return ans;
}
vector<long long> v[N];
map<long long, long long> cn;
void dfs(long long x, long long baap, long long l = 0) {
  cn[l]++;
  for (long long it : v[x]) {
    if (it != baap) dfs(it, x, l + 1);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  op();
  long long p = M;
  InverseofNumber(p);
  InverseofFactorial(p);
  factorial(p);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i <= n; i++) {
      v[i].clear();
    }
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    long long ans = 0;
    if (k == 2) {
      ans = (n * (n - 1) / 2) % M;
      cout << (ans) << "\n";
      continue;
    }
    for (long long i = 1; i <= n; i++) {
      cn.clear();
      dfs(i, i);
      for (pair<long long, long long> it : cn) {
        if (it.second >= k) {
          ans = (ans + nCr(it.second, k)) % M;
        }
      }
    }
    cout << (ans) << "\n";
  }
  return 0;
}
