#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m, t, sum, vis[150], x, y, it;
long long f[105], res, f2[105][105];
vector<int> v[205];
void dfs(int x, int fa, int d) {
  vis[x] = 1;
  if (d == j) {
    f[x]++;
    return;
  }
  for (auto i : v[x]) {
    if (i == fa) {
      continue;
    }
    dfs(i, x, d + 1);
    f[x] += f[i];
  }
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++) {
      v[i].clear();
    }
    for (i = 1; i < n; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    res = 0;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        memset(vis, 0, sizeof(vis));
        memset(f, 0, sizeof(f));
        memset(f2, 0, sizeof(f2));
        dfs(i, 0, 0);
        f2[0][0] = 1;
        it = 0;
        for (auto jj : v[i]) {
          if (f[jj] == 0) {
            continue;
          }
          for (k = 0; k <= it; k++) {
            f2[it][k] %= 1000000007;
            f2[it + 1][k] += f2[it][k];
            f2[it + 1][k + 1] += f2[it][k] * f[jj] % 1000000007;
          }
          it++;
        }
        res += f2[it][m];
        res %= 1000000007;
      }
    }
    printf("%lld\n", res);
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int i, j, k, n, m, t, sum, vis[150], x, y, it;
long long f[105], res, f2[105][105];
vector<int> v[205];
void dfs(int x, int fa, int d) {
  vis[x] = 1;
  if (d == j) {
    f[x]++;
    return;
  }
  for (auto i : v[x]) {
    if (i == fa) {
      continue;
    }
    dfs(i, x, d + 1);
    f[x] += f[i];
  }
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++) {
      v[i].clear();
    }
    for (i = 1; i < n; i++) {
      scanf("%d%d", &x, &y);
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    res = 0;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        memset(vis, 0, sizeof(vis));
        memset(f, 0, sizeof(f));
        memset(f2, 0, sizeof(f2));
        dfs(i, 0, 0);
        f2[0][0] = 1;
        it = 0;
        for (auto jj : v[i]) {
          if (f[jj] == 0) {
            continue;
          }
          for (k = 0; k <= it; k++) {
            f2[it][k] %= 1000000007;
            f2[it + 1][k] += f2[it][k];
            f2[it + 1][k + 1] += f2[it][k] * f[jj] % 1000000007;
          }
          it++;
        }
        res += f2[it][m];
        res %= 1000000007;
      }
    }
    printf("%lld\n", res);
  }
}

```