#include <bits/stdc++.h>
using namespace std;
const int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 105;
int n, k;
vector<int> adj[N];
int h[N];
void dfs(int u, int pre, int &cnt, int len) {
  if (h[u] == len) cnt++;
  for (int v : adj[u])
    if (v != pre) {
      h[v] = h[u] + 1;
      dfs(v, u, cnt, len);
    }
}
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void gogo() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int ans = 0;
  if (k == 2)
    cout << 1LL * n * (n - 1) / 2 << '\n';
  else {
    for (int len = 0; len <= n; ++len) {
      for (int i = 1; i <= n; ++i) {
        memset(h, 0, sizeof(h));
        vector<int> all;
        for (int u : adj[i]) {
          int cnt = 0;
          dfs(u, i, cnt, len);
          all.push_back(cnt);
        }
        vector<int> dp(k + 5);
        dp[0] = 1;
        for (int ptr = 0; ptr < (int)all.size(); ++ptr) {
          vector<int> f(k + 5, 0);
          for (int j = 0; j <= k; ++j) {
            if (dp[j] == 0) continue;
            add(f[j], dp[j]);
            add(f[j + 1], (long long)dp[j] * all[ptr] % mod);
          }
          swap(dp, f);
        }
        add(ans, dp[k]);
      }
    }
    cout << ans << '\n';
  }
  for (int i = 1; i <= n; ++i) adj[i].clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
}

### Prompt

Please create a solution in cpp 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 int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
const int dxx[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          dyy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long mod = 1000000007;
const int base = 311;
const int N = 105;
int n, k;
vector<int> adj[N];
int h[N];
void dfs(int u, int pre, int &cnt, int len) {
  if (h[u] == len) cnt++;
  for (int v : adj[u])
    if (v != pre) {
      h[v] = h[u] + 1;
      dfs(v, u, cnt, len);
    }
}
void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
void gogo() {
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int ans = 0;
  if (k == 2)
    cout << 1LL * n * (n - 1) / 2 << '\n';
  else {
    for (int len = 0; len <= n; ++len) {
      for (int i = 1; i <= n; ++i) {
        memset(h, 0, sizeof(h));
        vector<int> all;
        for (int u : adj[i]) {
          int cnt = 0;
          dfs(u, i, cnt, len);
          all.push_back(cnt);
        }
        vector<int> dp(k + 5);
        dp[0] = 1;
        for (int ptr = 0; ptr < (int)all.size(); ++ptr) {
          vector<int> f(k + 5, 0);
          for (int j = 0; j <= k; ++j) {
            if (dp[j] == 0) continue;
            add(f[j], dp[j]);
            add(f[j + 1], (long long)dp[j] * all[ptr] % mod);
          }
          swap(dp, f);
        }
        add(ans, dp[k]);
      }
    }
    cout << ans << '\n';
  }
  for (int i = 1; i <= n; ++i) adj[i].clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  if (fopen("sol"
            ".inp",
            "r")) {
    freopen(
        "sol"
        ".inp",
        "r", stdin);
    freopen(
        "sol"
        ".out",
        "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) gogo();
}

```