#include <bits/stdc++.h>
using namespace std;
long long nCrModp(int n, int r, int p) {
  if (r > n - r) r = n - r;
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
int n, k;
vector<vector<int> > graph(105);
vector<bool> visited(105);
long long NCR[105][105];
int dfs(int curr, int dist) {
  visited[curr] = true;
  if (dist == 0) return 1;
  int ans = 0;
  for (int a = 0; a < graph[curr].size(); a++) {
    int x = graph[curr][a];
    if (visited[x]) continue;
    ans += dfs(x, dist - 1);
  }
  return ans;
}
void solve() {
  cin >> n >> k;
  for (int a = 0; a <= n; a++) {
    graph[a].clear();
  }
  int x, y;
  for (int a = 1; a < n; a++) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  if (k == 2) {
    cout << NCR[n][2] << '\n';
    return;
  }
  long long ans = 0;
  long long pairs = 0;
  long long dp[k + 1][2];
  for (int a = 2; a < n; a += 2) {
    for (int b = 1; b <= n; b++) {
      if (graph[b].size() < k) {
        continue;
      }
      for (int c = 1; c <= n; c++) {
        visited[c] = false;
      }
      for (int c = 0; c <= k; c++) {
        dp[c][0] = 0;
        dp[c][1] = 0;
      }
      dp[0][1] = 1;
      dp[0][0] = 1;
      visited[b] = true;
      int cnt = 0, last, now;
      for (int c = 0; c < graph[b].size(); c++) {
        now = c % 2;
        last = (c + 1) % 2;
        for (int d = 0; d <= min(c + 1, k); d++) {
          dp[d][now] = dp[d][last];
        }
        cnt = dfs(graph[b][c], a / 2 - 1);
        if (cnt == 0) {
          continue;
        }
        for (int d = 1; d <= min(c + 1, k); d++) {
          dp[d][now] += (cnt * dp[d - 1][last]) % 1000000007;
          dp[d][now] %= 1000000007;
        }
      }
      ans += dp[k][now];
      ans %= 1000000007;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int a = 0; a < 105; a++) {
    for (int b = 0; b <= a; b++) {
      NCR[a][b] = nCrModp(a, b, 1000000007);
    }
  }
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
long long nCrModp(int n, int r, int p) {
  if (r > n - r) r = n - r;
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = min(i, r); j > 0; j--) C[j] = (C[j] + C[j - 1]) % p;
  }
  return C[r];
}
int n, k;
vector<vector<int> > graph(105);
vector<bool> visited(105);
long long NCR[105][105];
int dfs(int curr, int dist) {
  visited[curr] = true;
  if (dist == 0) return 1;
  int ans = 0;
  for (int a = 0; a < graph[curr].size(); a++) {
    int x = graph[curr][a];
    if (visited[x]) continue;
    ans += dfs(x, dist - 1);
  }
  return ans;
}
void solve() {
  cin >> n >> k;
  for (int a = 0; a <= n; a++) {
    graph[a].clear();
  }
  int x, y;
  for (int a = 1; a < n; a++) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  if (k == 2) {
    cout << NCR[n][2] << '\n';
    return;
  }
  long long ans = 0;
  long long pairs = 0;
  long long dp[k + 1][2];
  for (int a = 2; a < n; a += 2) {
    for (int b = 1; b <= n; b++) {
      if (graph[b].size() < k) {
        continue;
      }
      for (int c = 1; c <= n; c++) {
        visited[c] = false;
      }
      for (int c = 0; c <= k; c++) {
        dp[c][0] = 0;
        dp[c][1] = 0;
      }
      dp[0][1] = 1;
      dp[0][0] = 1;
      visited[b] = true;
      int cnt = 0, last, now;
      for (int c = 0; c < graph[b].size(); c++) {
        now = c % 2;
        last = (c + 1) % 2;
        for (int d = 0; d <= min(c + 1, k); d++) {
          dp[d][now] = dp[d][last];
        }
        cnt = dfs(graph[b][c], a / 2 - 1);
        if (cnt == 0) {
          continue;
        }
        for (int d = 1; d <= min(c + 1, k); d++) {
          dp[d][now] += (cnt * dp[d - 1][last]) % 1000000007;
          dp[d][now] %= 1000000007;
        }
      }
      ans += dp[k][now];
      ans %= 1000000007;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (int a = 0; a < 105; a++) {
    for (int b = 0; b <= a; b++) {
      NCR[a][b] = nCrModp(a, b, 1000000007);
    }
  }
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

```