#include <bits/stdc++.h>
using namespace std;
const int N = 105;
const int mod = 1000000007;
int dp[N][N], cnt[N];
vector<int> g[N];
void dfs(int v, int p, int d) {
  cnt[d]++;
  for (int u : g[v]) {
    if (u != p) dfs(u, v, d + 1);
  }
}
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
    for (int i = 2; i <= n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    int answer = 0;
    for (int root = 1; root <= n; root++) {
      if (g[root].size() < k) continue;
      memset(dp, 0, sizeof(dp));
      for (int d = 1; d <= n; d++) {
        dp[d][0] = 1;
      }
      for (int v : g[root]) {
        memset(cnt, 0, sizeof(cnt));
        dfs(v, root, 1);
        for (int d = 1; d <= n; d++) {
          for (int cur = k - 1; cur >= 0; cur--) {
            add(dp[d][cur + 1], dp[d][cur] * 1LL * cnt[d] % mod);
          }
        }
      }
      for (int d = 1; d <= n; d++) {
        add(answer, dp[d][k]);
      }
    }
    cout << answer << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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 = 105;
const int mod = 1000000007;
int dp[N][N], cnt[N];
vector<int> g[N];
void dfs(int v, int p, int d) {
  cnt[d]++;
  for (int u : g[v]) {
    if (u != p) dfs(u, v, d + 1);
  }
}
inline void add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
    for (int i = 2; i <= n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].emplace_back(v);
      g[v].emplace_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      continue;
    }
    int answer = 0;
    for (int root = 1; root <= n; root++) {
      if (g[root].size() < k) continue;
      memset(dp, 0, sizeof(dp));
      for (int d = 1; d <= n; d++) {
        dp[d][0] = 1;
      }
      for (int v : g[root]) {
        memset(cnt, 0, sizeof(cnt));
        dfs(v, root, 1);
        for (int d = 1; d <= n; d++) {
          for (int cur = k - 1; cur >= 0; cur--) {
            add(dp[d][cur + 1], dp[d][cur] * 1LL * cnt[d] % mod);
          }
        }
      }
      for (int d = 1; d <= n; d++) {
        add(answer, dp[d][k]);
      }
    }
    cout << answer << '\n';
  }
  return 0;
}

```