#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
inline void add(int &x, int &y) { (x += y) >= mod ? x -= mod : 0; }
inline int pl(int x, int y) { return (x += y) >= mod ? x -= mod : x; }
int T, n, m;
vector<int> E[110];
int C[110][110];
int cnt[110][110], tot, dp[110];
void getdis(int x, int fa, int dis) {
  ++cnt[tot][dis];
  for (auto y : E[x]) {
    if (y == fa) continue;
    getdis(y, x, dis + 1);
  }
}
int main() {
  C[0][0] = 1;
  for (int i = 1; i <= 100; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j) C[i][j] = pl(C[i - 1][j - 1], C[i - 1][j]);
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &m);
    for (int i = 1, x, y; i < n; ++i) {
      scanf("%d %d", &x, &y);
      E[x].emplace_back(y);
      E[y].emplace_back(x);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      for (int i = 1; i <= n; ++i) E[i].clear();
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (auto x : E[i]) ++tot, getdis(x, i, 1);
      for (int j = 1; j <= n; ++j) {
        int cnt1 = 0;
        for (int k = 1; k <= tot; ++k) cnt1 += cnt[k][j] != 0;
        if (cnt1 < m) break;
        dp[0] = 1;
        for (int k = 1; k <= tot; ++k)
          for (int l = k; l; --l)
            dp[l] = (dp[l] + 1ll * dp[l - 1] * cnt[k][j]) % mod;
        add(ans, dp[m]);
        memset(dp, 0, tot + 1 << 2);
      }
      for (int j = 1; j <= tot; ++j) memset(cnt[j] + 1, 0, n << 2);
      tot = 0;
    }
    printf("%d\n", ans);
    for (int i = 1; i <= n; ++i) E[i].clear();
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 mod = 1000000007;
inline void add(int &x, int &y) { (x += y) >= mod ? x -= mod : 0; }
inline int pl(int x, int y) { return (x += y) >= mod ? x -= mod : x; }
int T, n, m;
vector<int> E[110];
int C[110][110];
int cnt[110][110], tot, dp[110];
void getdis(int x, int fa, int dis) {
  ++cnt[tot][dis];
  for (auto y : E[x]) {
    if (y == fa) continue;
    getdis(y, x, dis + 1);
  }
}
int main() {
  C[0][0] = 1;
  for (int i = 1; i <= 100; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j) C[i][j] = pl(C[i - 1][j - 1], C[i - 1][j]);
  }
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &m);
    for (int i = 1, x, y; i < n; ++i) {
      scanf("%d %d", &x, &y);
      E[x].emplace_back(y);
      E[y].emplace_back(x);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      for (int i = 1; i <= n; ++i) E[i].clear();
      continue;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      for (auto x : E[i]) ++tot, getdis(x, i, 1);
      for (int j = 1; j <= n; ++j) {
        int cnt1 = 0;
        for (int k = 1; k <= tot; ++k) cnt1 += cnt[k][j] != 0;
        if (cnt1 < m) break;
        dp[0] = 1;
        for (int k = 1; k <= tot; ++k)
          for (int l = k; l; --l)
            dp[l] = (dp[l] + 1ll * dp[l - 1] * cnt[k][j]) % mod;
        add(ans, dp[m]);
        memset(dp, 0, tot + 1 << 2);
      }
      for (int j = 1; j <= tot; ++j) memset(cnt[j] + 1, 0, n << 2);
      tot = 0;
    }
    printf("%d\n", ans);
    for (int i = 1; i <= n; ++i) E[i].clear();
  }
  return 0;
}

```