#include <bits/stdc++.h>
using namespace std;
const long long maxn = 109;
const long long mod = 1e9 + 7;
vector<long long> vec[maxn];
long long deep[maxn], zi[maxn][maxn], n, k;
void dfs(long long u, long long father, long long rt) {
  deep[u] = deep[father] + 1;
  zi[rt][deep[u]]++;
  for (auto v : vec[u]) {
    if (v == father) continue;
    dfs(v, u, rt);
  }
}
long long f[maxn][maxn], w[maxn];
void upd(long long &x, long long y) { x = (x + y) % mod; }
long long DP() {
  for (long long i = 1; i <= w[0]; i++)
    for (long long j = 0; j <= i && j <= k; j++) f[i][j] = 0;
  f[0][0] = 1;
  for (long long i = 1; i <= w[0]; i++)
    for (long long j = 0; j <= i && j <= k; j++) {
      if (j) upd(f[i][j], f[i - 1][j - 1] * w[i] % mod);
      upd(f[i][j], f[i - 1][j]);
    }
  return f[w[0]][k];
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (long long i = 1; i < n; i++) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      vec[l].push_back(r);
      vec[r].push_back(l);
    }
    if (k == 2)
      cout << n * (n - 1) / 2 << endl;
    else {
      long long res = 0;
      for (long long i = 1; i <= n; i++) {
        deep[i] = 0;
        for (auto v : vec[i]) dfs(v, i, v);
        for (long long j = 1; j <= n; j++) {
          w[0] = 0;
          for (auto v : vec[i]) w[++w[0]] = zi[v][j];
          res = (res + DP()) % mod;
        }
        for (auto v : vec[i])
          for (long long j = 1; j <= n; j++) zi[v][j] = 0;
      }
      cout << res << endl;
    }
    for (long long i = 1; i <= n; i++) vec[i].clear();
  }
}

### 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;
const long long maxn = 109;
const long long mod = 1e9 + 7;
vector<long long> vec[maxn];
long long deep[maxn], zi[maxn][maxn], n, k;
void dfs(long long u, long long father, long long rt) {
  deep[u] = deep[father] + 1;
  zi[rt][deep[u]]++;
  for (auto v : vec[u]) {
    if (v == father) continue;
    dfs(v, u, rt);
  }
}
long long f[maxn][maxn], w[maxn];
void upd(long long &x, long long y) { x = (x + y) % mod; }
long long DP() {
  for (long long i = 1; i <= w[0]; i++)
    for (long long j = 0; j <= i && j <= k; j++) f[i][j] = 0;
  f[0][0] = 1;
  for (long long i = 1; i <= w[0]; i++)
    for (long long j = 0; j <= i && j <= k; j++) {
      if (j) upd(f[i][j], f[i - 1][j - 1] * w[i] % mod);
      upd(f[i][j], f[i - 1][j]);
    }
  return f[w[0]][k];
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (long long i = 1; i < n; i++) {
      long long l, r;
      scanf("%lld%lld", &l, &r);
      vec[l].push_back(r);
      vec[r].push_back(l);
    }
    if (k == 2)
      cout << n * (n - 1) / 2 << endl;
    else {
      long long res = 0;
      for (long long i = 1; i <= n; i++) {
        deep[i] = 0;
        for (auto v : vec[i]) dfs(v, i, v);
        for (long long j = 1; j <= n; j++) {
          w[0] = 0;
          for (auto v : vec[i]) w[++w[0]] = zi[v][j];
          res = (res + DP()) % mod;
        }
        for (auto v : vec[i])
          for (long long j = 1; j <= n; j++) zi[v][j] = 0;
      }
      cout << res << endl;
    }
    for (long long i = 1; i <= n; i++) vec[i].clear();
  }
}

```