#include <bits/stdc++.h>
using namespace std;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long N = 1e2 + 5;
long long freq[N], n, k;
vector<long long> v[N];
void dfs(long long z, long long p, long long d = 1) {
  freq[d]++;
  for (long long x : v[z])
    if (x ^ p) dfs(x, z, d + 1);
}
void test_case() {
  cin >> n >> k;
  for (long long i = 1; i < n + 1; i++) v[i].clear();
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y), v[y].push_back(x);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    return;
  }
  long long ans = 0;
  for (long long center = 1; center < n + 1; center++) {
    vector<vector<long long>> cnt(n + 1);
    for (long long x : v[center]) {
      memset(freq, 0, sizeof freq);
      dfs(x, center);
      for (long long j = 0; j < n; j++) {
        if (freq[j]) cnt[j].push_back(freq[j]);
      }
    }
    for (long long dep = 0; dep < n; dep++) {
      vector<long long> dp(k + 1);
      dp[0] = 1;
      for (long long x : cnt[dep]) {
        for (long long j = k - 1; j >= 0; j--) {
          dp[j + 1] = (dp[j + 1] + dp[j] * x) % 1000000007;
        }
      }
      ans = (ans + dp[k]) % 1000000007;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    test_case();
  }
  return 0;
}

### 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;
template <typename T, typename T1>
T amax(T &a, T1 b) {
  if (b > a) a = b;
  return a;
}
template <typename T, typename T1>
T amin(T &a, T1 b) {
  if (b < a) a = b;
  return a;
}
const long long N = 1e2 + 5;
long long freq[N], n, k;
vector<long long> v[N];
void dfs(long long z, long long p, long long d = 1) {
  freq[d]++;
  for (long long x : v[z])
    if (x ^ p) dfs(x, z, d + 1);
}
void test_case() {
  cin >> n >> k;
  for (long long i = 1; i < n + 1; i++) v[i].clear();
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y), v[y].push_back(x);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << "\n";
    return;
  }
  long long ans = 0;
  for (long long center = 1; center < n + 1; center++) {
    vector<vector<long long>> cnt(n + 1);
    for (long long x : v[center]) {
      memset(freq, 0, sizeof freq);
      dfs(x, center);
      for (long long j = 0; j < n; j++) {
        if (freq[j]) cnt[j].push_back(freq[j]);
      }
    }
    for (long long dep = 0; dep < n; dep++) {
      vector<long long> dp(k + 1);
      dp[0] = 1;
      for (long long x : cnt[dep]) {
        for (long long j = k - 1; j >= 0; j--) {
          dp[j + 1] = (dp[j + 1] + dp[j] * x) % 1000000007;
        }
      }
      ans = (ans + dp[k]) % 1000000007;
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    test_case();
  }
  return 0;
}

```