#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long int mod = 1e9 + 7;
const int M = 105;
long long int n, k;
vector<long long int> a[M];
long long int d[M][M];
long long int child;
void dfs(long long int v, long long int p, long long int dep) {
  d[child][dep]++;
  for (long long int u : a[v]) {
    if (u == p) continue;
    dfs(u, v, dep + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TT = clock();
  long long int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (long long int i = 0; i < n; i++) a[i].clear();
    for (long long int i = 0; i < n - 1; i++) {
      long long int u, v;
      cin >> u >> v;
      u--, v--;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    if (k == 2) {
      long long int ans = n * (n - 1) / 2;
      ans %= mod;
      cout << ans << "\n";
      continue;
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      memset(d, 0, sizeof(d));
      for (long long int j = 0; j < a[i].size(); j++) {
        child = j;
        dfs(a[i][j], i, 1);
      }
      for (long long int dep = 1; dep <= n; dep++) {
        vector<long long int> c;
        for (long long int j = 0; j < n; j++)
          if (d[j][dep] != 0) c.push_back(d[j][dep]);
        long long int r = c.size();
        if (r < k) continue;
        vector<vector<long long int>> dp(r + 1, vector<long long int>(k + 2));
        dp[r][k] = 1;
        for (long long int i = r - 1; i >= 0; i--) {
          for (long long int j = k; j >= 0; j--) {
            dp[i][j] = (dp[i + 1][j] + dp[i + 1][j + 1] * c[i] % mod) % mod;
          }
        }
        ans = (ans + dp[0][0]) % mod;
      }
    }
    cout << ans << "\n";
  }
  cerr << "\n\nTIME: " << (long double)(clock() - TT) / CLOCKS_PER_SEC
       << " sec\n";
  TT = clock();
  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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long int mod = 1e9 + 7;
const int M = 105;
long long int n, k;
vector<long long int> a[M];
long long int d[M][M];
long long int child;
void dfs(long long int v, long long int p, long long int dep) {
  d[child][dep]++;
  for (long long int u : a[v]) {
    if (u == p) continue;
    dfs(u, v, dep + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TT = clock();
  long long int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (long long int i = 0; i < n; i++) a[i].clear();
    for (long long int i = 0; i < n - 1; i++) {
      long long int u, v;
      cin >> u >> v;
      u--, v--;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    if (k == 2) {
      long long int ans = n * (n - 1) / 2;
      ans %= mod;
      cout << ans << "\n";
      continue;
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      memset(d, 0, sizeof(d));
      for (long long int j = 0; j < a[i].size(); j++) {
        child = j;
        dfs(a[i][j], i, 1);
      }
      for (long long int dep = 1; dep <= n; dep++) {
        vector<long long int> c;
        for (long long int j = 0; j < n; j++)
          if (d[j][dep] != 0) c.push_back(d[j][dep]);
        long long int r = c.size();
        if (r < k) continue;
        vector<vector<long long int>> dp(r + 1, vector<long long int>(k + 2));
        dp[r][k] = 1;
        for (long long int i = r - 1; i >= 0; i--) {
          for (long long int j = k; j >= 0; j--) {
            dp[i][j] = (dp[i + 1][j] + dp[i + 1][j + 1] * c[i] % mod) % mod;
          }
        }
        ans = (ans + dp[0][0]) % mod;
      }
    }
    cout << ans << "\n";
  }
  cerr << "\n\nTIME: " << (long double)(clock() - TT) / CLOCKS_PER_SEC
       << " sec\n";
  TT = clock();
  return 0;
}

```