#include <bits/stdc++.h>
using namespace std;
const char nl = '\n';
const long long inf = 1e9 + 10;
const long long inf2 = 1e18 + 99LL;
const long double inf3 = 1e17;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-9;
const bool local = false;
const int logn = 17, maxn = 101, maxm = 200001, maxn2 = 3;
vector<int> g[maxn], d, a[maxn];
void dfs(int v, int p, int cd = 0) {
  d[cd]++;
  for (int x : g[v]) {
    if (x == p) continue;
    dfs(x, v, cd + 1);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 0; i < n - 1; i++) {
    int ta, b;
    cin >> ta >> b;
    ta--;
    b--;
    g[ta].push_back(b);
    g[b].push_back(ta);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << nl;
    return;
  }
  long long ans = 0;
  for (int v = 0; v < n; v++) {
    if ((int)((g[v]).size()) < k) continue;
    for (int i = 0; i < n; i++) a[i].clear();
    for (int x : g[v]) {
      d.assign(n, 0);
      dfs(x, v);
      for (int j = 0; j < n; j++) {
        if (d[j] == 0) break;
        a[j].push_back(d[j]);
      }
    }
    for (int i = 0; i < n; i++) {
      if ((int)((a[i]).size()) < k) break;
      vector<vector<long long>> dp((int)((a[i]).size()) + 1,
                                   vector<long long>(k + 1));
      dp[0][0] = 1;
      for (int j = 1; j <= (int)((a[i]).size()); j++) {
        dp[j][0] = dp[j - 1][0];
        for (int tk = 1; tk <= k; tk++) {
          dp[j][tk] = dp[j - 1][tk - 1] * a[i][j - 1] + dp[j - 1][tk];
          dp[j][tk] %= mod;
        }
      }
      ans += dp.back().back();
      ans %= mod;
    }
  }
  cout << ans << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Create a solution in Cpp for 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 char nl = '\n';
const long long inf = 1e9 + 10;
const long long inf2 = 1e18 + 99LL;
const long double inf3 = 1e17;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-9;
const bool local = false;
const int logn = 17, maxn = 101, maxm = 200001, maxn2 = 3;
vector<int> g[maxn], d, a[maxn];
void dfs(int v, int p, int cd = 0) {
  d[cd]++;
  for (int x : g[v]) {
    if (x == p) continue;
    dfs(x, v, cd + 1);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) g[i].clear();
  for (int i = 0; i < n - 1; i++) {
    int ta, b;
    cin >> ta >> b;
    ta--;
    b--;
    g[ta].push_back(b);
    g[b].push_back(ta);
  }
  if (k == 2) {
    cout << n * (n - 1) / 2 << nl;
    return;
  }
  long long ans = 0;
  for (int v = 0; v < n; v++) {
    if ((int)((g[v]).size()) < k) continue;
    for (int i = 0; i < n; i++) a[i].clear();
    for (int x : g[v]) {
      d.assign(n, 0);
      dfs(x, v);
      for (int j = 0; j < n; j++) {
        if (d[j] == 0) break;
        a[j].push_back(d[j]);
      }
    }
    for (int i = 0; i < n; i++) {
      if ((int)((a[i]).size()) < k) break;
      vector<vector<long long>> dp((int)((a[i]).size()) + 1,
                                   vector<long long>(k + 1));
      dp[0][0] = 1;
      for (int j = 1; j <= (int)((a[i]).size()); j++) {
        dp[j][0] = dp[j - 1][0];
        for (int tk = 1; tk <= k; tk++) {
          dp[j][tk] = dp[j - 1][tk - 1] * a[i][j - 1] + dp[j - 1][tk];
          dp[j][tk] %= mod;
        }
      }
      ans += dp.back().back();
      ans %= mod;
    }
  }
  cout << ans << nl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) solve();
}

```