#include <bits/stdc++.h>
using namespace std;
const long long N = 2e2 + 10;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const long long INF = 0x3f3f3f3f;
long long n, k;
long long ne[N], h[N], e[N], idx = 1;
long long dp[N][N], cnt[N][N];
void add(long long a, long long b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
void dfs(long long u, long long fa) {
  cnt[u][0] = 1;
  for (long long i = h[u]; i; i = ne[i]) {
    long long j = e[i];
    if (j == fa) continue;
    dfs(j, u);
    for (long long deep = 1; deep <= n; deep++)
      cnt[u][deep] += cnt[j][deep - 1];
  }
}
long long slove(long long u) {
  memset(cnt, 0, sizeof(cnt));
  dfs(u, -1);
  vector<long long> son;
  son.push_back(-1);
  for (long long i = h[u]; i; i = ne[i]) son.push_back(e[i]);
  long long len = (long long)son.size(), res = 0;
  for (long long d = 0; d <= n; d++) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (long long i = 1; i < len; i++) {
      dp[i][0] = 1;
      for (long long j = 1; j <= i; j++) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
        dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * cnt[son[i]][d] % mod)) % mod;
      }
    }
    res = (res + dp[len - 1][k]) % mod;
  }
  return res;
}
signed main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    for (long long i = 1; i <= idx; i++) h[i] = 0;
    idx = 1;
    cin >> n >> k;
    for (long long i = 1; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      add(a, b);
      add(b, a);
    }
    long long ans = 0;
    if (k == 2)
      ans = (n * (n - 1) / 2) % mod;
    else
      for (long long root = 1; root <= n; root++)
        ans = (ans + slove(root)) % mod;
    cout << ans << endl;
  }
  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 long long N = 2e2 + 10;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const long long INF = 0x3f3f3f3f;
long long n, k;
long long ne[N], h[N], e[N], idx = 1;
long long dp[N][N], cnt[N][N];
void add(long long a, long long b) { e[idx] = b, ne[idx] = h[a], h[a] = idx++; }
void dfs(long long u, long long fa) {
  cnt[u][0] = 1;
  for (long long i = h[u]; i; i = ne[i]) {
    long long j = e[i];
    if (j == fa) continue;
    dfs(j, u);
    for (long long deep = 1; deep <= n; deep++)
      cnt[u][deep] += cnt[j][deep - 1];
  }
}
long long slove(long long u) {
  memset(cnt, 0, sizeof(cnt));
  dfs(u, -1);
  vector<long long> son;
  son.push_back(-1);
  for (long long i = h[u]; i; i = ne[i]) son.push_back(e[i]);
  long long len = (long long)son.size(), res = 0;
  for (long long d = 0; d <= n; d++) {
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (long long i = 1; i < len; i++) {
      dp[i][0] = 1;
      for (long long j = 1; j <= i; j++) {
        dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
        dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * cnt[son[i]][d] % mod)) % mod;
      }
    }
    res = (res + dp[len - 1][k]) % mod;
  }
  return res;
}
signed main() {
  long long tt;
  cin >> tt;
  while (tt--) {
    for (long long i = 1; i <= idx; i++) h[i] = 0;
    idx = 1;
    cin >> n >> k;
    for (long long i = 1; i < n; i++) {
      long long a, b;
      cin >> a >> b;
      add(a, b);
      add(b, a);
    }
    long long ans = 0;
    if (k == 2)
      ans = (n * (n - 1) / 2) % mod;
    else
      for (long long root = 1; root <= n; root++)
        ans = (ans + slove(root)) % mod;
    cout << ans << endl;
  }
  return 0;
}

```