#include <bits/stdc++.h>
using namespace std;
constexpr int64_t mod = 1e9 + 7;
constexpr int64_t mxn = 101;
int64_t d[mxn][mxn];
vector<int64_t> adj[mxn];
void dfs(int64_t u, int64_t p, int64_t l, int64_t r) {
  d[r][l]++;
  for (auto& v : adj[u]) {
    if (v != p) {
      dfs(v, u, l + 1, r);
    }
  }
}
int64_t dp[mxn][mxn];
int64_t compute(vector<int64_t>& f, int64_t k) {
  int64_t m = f.size();
  if (m < k) return 0;
  memset(dp, 0, sizeof dp);
  dp[0][0] = 1;
  dp[0][1] = f[0];
  for (int64_t i = 1; i < m; ++i) {
    dp[i][0] = 1;
    for (int64_t j = 1; j <= k; ++j) {
      dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * f[i] % mod) % mod;
    }
  }
  return dp[m - 1][k];
}
void hike() {
  int64_t n, k;
  cin >> n >> k;
  for (int64_t i = 0; i + 1 < n; ++i) {
    int64_t u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    for (int64_t i = 1; i <= n; ++i) {
      adj[i].clear();
    }
    return;
  }
  int64_t count = 0;
  for (int64_t i = 1; i <= n; ++i) {
    for (auto& v : adj[i]) {
      dfs(v, i, 1, v);
    }
    for (int64_t l = 1; l < n; ++l) {
      vector<int64_t> f;
      for (int64_t j = 1; j <= n; ++j) {
        if (d[j][l]) {
          f.push_back(d[j][l]);
        }
      }
      (count += compute(f, k)) %= mod;
    }
    for (int64_t i = 0; i <= n; ++i) {
      for (int64_t j = 0; j <= n; ++j) {
        d[i][j] = 0;
      }
    }
  }
  cout << count << "\n";
  for (int64_t i = 1; i <= n; ++i) {
    adj[i].clear();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int64_t t = 1;
  cin >> t;
  while (t--) hike();
  return 0;
}

### Prompt

Create a solution in CPP for 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;
constexpr int64_t mod = 1e9 + 7;
constexpr int64_t mxn = 101;
int64_t d[mxn][mxn];
vector<int64_t> adj[mxn];
void dfs(int64_t u, int64_t p, int64_t l, int64_t r) {
  d[r][l]++;
  for (auto& v : adj[u]) {
    if (v != p) {
      dfs(v, u, l + 1, r);
    }
  }
}
int64_t dp[mxn][mxn];
int64_t compute(vector<int64_t>& f, int64_t k) {
  int64_t m = f.size();
  if (m < k) return 0;
  memset(dp, 0, sizeof dp);
  dp[0][0] = 1;
  dp[0][1] = f[0];
  for (int64_t i = 1; i < m; ++i) {
    dp[i][0] = 1;
    for (int64_t j = 1; j <= k; ++j) {
      dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * f[i] % mod) % mod;
    }
  }
  return dp[m - 1][k];
}
void hike() {
  int64_t n, k;
  cin >> n >> k;
  for (int64_t i = 0; i + 1 < n; ++i) {
    int64_t u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    for (int64_t i = 1; i <= n; ++i) {
      adj[i].clear();
    }
    return;
  }
  int64_t count = 0;
  for (int64_t i = 1; i <= n; ++i) {
    for (auto& v : adj[i]) {
      dfs(v, i, 1, v);
    }
    for (int64_t l = 1; l < n; ++l) {
      vector<int64_t> f;
      for (int64_t j = 1; j <= n; ++j) {
        if (d[j][l]) {
          f.push_back(d[j][l]);
        }
      }
      (count += compute(f, k)) %= mod;
    }
    for (int64_t i = 0; i <= n; ++i) {
      for (int64_t j = 0; j <= n; ++j) {
        d[i][j] = 0;
      }
    }
  }
  cout << count << "\n";
  for (int64_t i = 1; i <= n; ++i) {
    adj[i].clear();
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int64_t t = 1;
  cin >> t;
  while (t--) hike();
  return 0;
}

```