#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > ed;
const int m = 1e9 + 7;
void dfs(int cur, int pr, vector<int>& dist) {
  dist[cur] = dist[pr] + 1;
  for (auto to : ed[cur]) {
    if (to == pr) continue;
    dfs(to, cur, dist);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  ed.clear();
  ed.resize(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    return;
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    vector<int> dist(n + 1);
    vector<vector<int> > arr(n + 1, vector<int>(n + 1));
    for (auto to : ed[i]) {
      dfs(to, i, dist);
      for (int j = 1; j <= n; ++j) {
        if (dist[j] == 0) continue;
        arr[to][dist[j]]++;
      }
      dist.assign(n + 1, 0);
    }
    for (int l = 1; l <= n; ++l) {
      vector<vector<int> > dp(n + 1, vector<int>(n + 1));
      int prev = 0;
      dp[0][0] = 1;
      for (auto to : ed[i]) {
        for (int j = 0; j <= n; ++j) {
          dp[to][j] = dp[prev][j];
          if (j != 0)
            dp[to][j] += (long long)dp[prev][j - 1] * arr[to][l] % m,
                dp[to][j] %= m;
        }
        prev = to;
      }
      ans += dp[prev][k];
      ans %= m;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tst;
  tst = 1;
  cin >> tst;
  while (tst--) {
    solve();
  }
  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;
vector<vector<int> > ed;
const int m = 1e9 + 7;
void dfs(int cur, int pr, vector<int>& dist) {
  dist[cur] = dist[pr] + 1;
  for (auto to : ed[cur]) {
    if (to == pr) continue;
    dfs(to, cur, dist);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  ed.clear();
  ed.resize(n + 1);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    return;
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    vector<int> dist(n + 1);
    vector<vector<int> > arr(n + 1, vector<int>(n + 1));
    for (auto to : ed[i]) {
      dfs(to, i, dist);
      for (int j = 1; j <= n; ++j) {
        if (dist[j] == 0) continue;
        arr[to][dist[j]]++;
      }
      dist.assign(n + 1, 0);
    }
    for (int l = 1; l <= n; ++l) {
      vector<vector<int> > dp(n + 1, vector<int>(n + 1));
      int prev = 0;
      dp[0][0] = 1;
      for (auto to : ed[i]) {
        for (int j = 0; j <= n; ++j) {
          dp[to][j] = dp[prev][j];
          if (j != 0)
            dp[to][j] += (long long)dp[prev][j - 1] * arr[to][l] % m,
                dp[to][j] %= m;
        }
        prev = to;
      }
      ans += dp[prev][k];
      ans %= m;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tst;
  tst = 1;
  cin >> tst;
  while (tst--) {
    solve();
  }
  return 0;
}

```