#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
vector<T> make_unique(vector<T> a) {
  sort((a).begin(), (a).end());
  a.erase(unique((a).begin(), (a).end()), a.end());
  return a;
}
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int N = 111;
int q[N], dist[N], root[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> adj(n);
    for (int _ = 0; _ < (int)(n - 1); _++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      cout << (long long int)n * (n - 1) / 2 << endl;
      continue;
    }
    long long int ans = 0;
    for (int s = 0; s < (int)(n); s++) {
      fill_n(dist, n, INF);
      int l = 0, r = 0;
      dist[s] = 0;
      q[r++] = s;
      while (l < r) {
        int u = q[l++];
        for (auto v : adj[u]) {
          if (dist[v] > dist[u] + 1) {
            dist[v] = dist[u] + 1;
            root[v] = (u == s ? v : root[u]);
            q[r++] = v;
          }
        }
      }
      vector<map<int, int>> cnt(n);
      for (int u = 0; u < (int)(n); u++) cnt[dist[u]][root[u]]++;
      for (int d = 0; d < (int)(n); d++) {
        vector<int> poly(n + 1);
        poly[0] = 1;
        for (auto x : cnt[d]) {
          for (int i = (int)(n)-1; i >= ((int)0); i--) {
            poly[i + 1] += (long long int)x.second * poly[i] % MOD;
            if (poly[i + 1] >= MOD) poly[i + 1] -= MOD;
          }
        }
        ans += poly[k];
      }
    }
    cout << ans % MOD << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
vector<T> make_unique(vector<T> a) {
  sort((a).begin(), (a).end());
  a.erase(unique((a).begin(), (a).end()), a.end());
  return a;
}
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int N = 111;
int q[N], dist[N], root[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> adj(n);
    for (int _ = 0; _ < (int)(n - 1); _++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      cout << (long long int)n * (n - 1) / 2 << endl;
      continue;
    }
    long long int ans = 0;
    for (int s = 0; s < (int)(n); s++) {
      fill_n(dist, n, INF);
      int l = 0, r = 0;
      dist[s] = 0;
      q[r++] = s;
      while (l < r) {
        int u = q[l++];
        for (auto v : adj[u]) {
          if (dist[v] > dist[u] + 1) {
            dist[v] = dist[u] + 1;
            root[v] = (u == s ? v : root[u]);
            q[r++] = v;
          }
        }
      }
      vector<map<int, int>> cnt(n);
      for (int u = 0; u < (int)(n); u++) cnt[dist[u]][root[u]]++;
      for (int d = 0; d < (int)(n); d++) {
        vector<int> poly(n + 1);
        poly[0] = 1;
        for (auto x : cnt[d]) {
          for (int i = (int)(n)-1; i >= ((int)0); i--) {
            poly[i + 1] += (long long int)x.second * poly[i] % MOD;
            if (poly[i + 1] >= MOD) poly[i + 1] -= MOD;
          }
        }
        ans += poly[k];
      }
    }
    cout << ans % MOD << endl;
  }
  return 0;
}

```