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>
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define ALL(a) begin(a), end(a)
#define RALL(a) rbegin(a), rend(a)
#define REP(i, a) for (int i(0); i < int(a); ++i)
#define FOR(i, a, b) for (int i(a); i <= int(b); ++i)

#ifdef LOCAL
#include <mydebug.h>
#else
#define debug(...)
#endif

using namespace std;
using ll = long long;

const ll MOD = 1e9 + 7;
const ll N = 100;
ll fact[N + 5] = {1, 1};
ll inv[N + 5] = {0, 1};
ll invfact[N + 5] = {1, 1};

void init() {
  FOR(i, 2, N) {
    fact[i] = (fact[i - 1] * i) % MOD;
    inv[i] = MOD - MOD / i * inv[MOD % i] % MOD;
    invfact[i] = invfact[i - 1] * inv[i] % MOD;
  }
}

ll nCk(ll n, ll k) { return fact[n] * invfact[k] % MOD * invfact[n - k] % MOD; }

void solve() {
  int n, k;
  cin >> n >> k;

  vector<vector<int>> adj(n);
  REP(i, n - 1) {
    int u, v;
    cin >> u >> v, --u, --v;
    adj[u].EB(v);
    adj[v].EB(u);
  }

  auto run = [&](int root) -> vector<int> {
    vector<int> dist(n);
    auto dfs = [&](auto &f, int u, int par, int dep) -> void {
      ++dist[dep];
      for (int v : adj[u]) {
        if (v == par) {
          continue;
        }
        f(f, v, u, dep + 1);
      }
    };
    dfs(dfs, root, -1, 0);
    return dist;
  };

  ll ans = 0;

  REP(root, n) {
    auto dist = run(root);
    debug(root, dist);
    for (int sz : dist) {
      if (sz < k) {
        continue;
      }
      ans = (ans + nCk(sz, k)) % MOD;
    }
  }
  cout << ans << "\n";
}

int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);

  init();

  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}
