#include <bits/stdc++.h>
using namespace std;
const int N = 110;
const long long mod = 1e9 + 7LL;
vector<int> lista[N];
int vis[N], g[N], peso[N], qt;
long long dp[N][N], visdp[N][N], t = 0;
long long solve(int pos, int k) {
  if (!k) return 1;
  if (pos == qt) return 0;
  long long &ans = dp[pos][k];
  if (visdp[pos][k] == t) return ans;
  visdp[pos][k] = t;
  ans = (peso[pos] * solve(pos + 1, k - 1) + solve(pos + 1, k)) % mod;
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, c;
  int q;
  cin >> q;
  while (q--) {
    cin >> n >> c;
    for (int i = 0; i <= n; ++i) {
      vis[i] = 0;
      lista[i].clear();
    }
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      lista[a].push_back(b);
      lista[b].push_back(a);
    }
    long long ans = 0;
    if (c == 2) {
      cout << n * (n - 1) / 2 << "\n";
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      queue<int> f, f1;
      vis[i] = i;
      qt = 0;
      for (int u : lista[i]) {
        g[u] = qt++;
        peso[g[u]] = 1;
        vis[u] = i;
        f.push(u);
      }
      t++;
      ans = (ans + solve(0, c)) % mod;
      while (!f.empty()) {
        while (!f.empty()) {
          int u = f.front();
          f.pop();
          peso[g[u]]--;
          for (int v : lista[u])
            if (vis[v] != i) {
              g[v] = g[u];
              peso[g[v]]++;
              vis[v] = i;
              f1.push(v);
            }
        }
        t++;
        ans = (ans + solve(0, c)) % mod;
        swap(f1, f);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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;
const int N = 110;
const long long mod = 1e9 + 7LL;
vector<int> lista[N];
int vis[N], g[N], peso[N], qt;
long long dp[N][N], visdp[N][N], t = 0;
long long solve(int pos, int k) {
  if (!k) return 1;
  if (pos == qt) return 0;
  long long &ans = dp[pos][k];
  if (visdp[pos][k] == t) return ans;
  visdp[pos][k] = t;
  ans = (peso[pos] * solve(pos + 1, k - 1) + solve(pos + 1, k)) % mod;
  return ans;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, c;
  int q;
  cin >> q;
  while (q--) {
    cin >> n >> c;
    for (int i = 0; i <= n; ++i) {
      vis[i] = 0;
      lista[i].clear();
    }
    for (int i = 0; i < n - 1; ++i) {
      int a, b;
      cin >> a >> b;
      lista[a].push_back(b);
      lista[b].push_back(a);
    }
    long long ans = 0;
    if (c == 2) {
      cout << n * (n - 1) / 2 << "\n";
      continue;
    }
    for (int i = 1; i <= n; ++i) {
      queue<int> f, f1;
      vis[i] = i;
      qt = 0;
      for (int u : lista[i]) {
        g[u] = qt++;
        peso[g[u]] = 1;
        vis[u] = i;
        f.push(u);
      }
      t++;
      ans = (ans + solve(0, c)) % mod;
      while (!f.empty()) {
        while (!f.empty()) {
          int u = f.front();
          f.pop();
          peso[g[u]]--;
          for (int v : lista[u])
            if (vis[v] != i) {
              g[v] = g[u];
              peso[g[v]]++;
              vis[v] = i;
              f1.push(v);
            }
        }
        t++;
        ans = (ans + solve(0, c)) % mod;
        swap(f1, f);
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```