#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const LL mod = 1e9 + 7;
int n, k, u, v;
vector<int> graph[205];
LL dp[205][205];
bool vis[205];
LL solve(int m, int k, vector<int> &cnt) {
  dp[0][0] = 1;
  for (int j = 1; j <= m; ++j) {
    for (int l = 0; l <= k; ++l) {
      dp[j][l] = (dp[j - 1][l] + dp[j][l]) % mod;
      if (l > 0) {
        dp[j][l] = (dp[j - 1][l - 1] * cnt[j - 1] % mod + dp[j][l]) % mod;
      }
    }
  }
  LL ans = dp[m][k];
  for (int j = 0; j <= m; ++j) {
    for (int l = 0; l <= k; ++l) {
      dp[j][l] = 0;
    }
  }
  fill_n(cnt.begin(), m, 0);
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    for (int i = 1; i < n; ++i) {
      scanf("%d%d", &u, &v);
      graph[u].push_back(v);
      graph[v].push_back(u);
    }
    if (k > 2) {
      LL ans = 0;
      for (int i = 1; i <= n; ++i) {
        fill_n(vis + 1, n, false);
        vis[i] = true;
        vector<int> cnt(graph[i].size(), 0);
        queue<tuple<int, int, int>> quu;
        for (int j = 0; j < graph[i].size(); ++j) {
          quu.push({graph[i][j], j, 1});
        }
        int nowHeight = 1;
        while (!quu.empty()) {
          auto cur = quu.front();
          quu.pop();
          vis[get<0>(cur)] = true;
          if (get<2>(cur) > nowHeight) {
            ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
            ++nowHeight;
          }
          ++cnt[get<1>(cur)];
          for (auto nxt : graph[get<0>(cur)]) {
            if (vis[nxt]) continue;
            quu.push({nxt, get<1>(cur), get<2>(cur) + 1});
          }
        }
        ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
      }
      printf("%lld\n", ans);
    } else {
      printf("%lld\n", 1ll * n * (n - 1) / 2 % mod);
    }
    for (int i = 1; i <= n; ++i) {
      graph[i].clear();
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
using LL = long long;
const LL mod = 1e9 + 7;
int n, k, u, v;
vector<int> graph[205];
LL dp[205][205];
bool vis[205];
LL solve(int m, int k, vector<int> &cnt) {
  dp[0][0] = 1;
  for (int j = 1; j <= m; ++j) {
    for (int l = 0; l <= k; ++l) {
      dp[j][l] = (dp[j - 1][l] + dp[j][l]) % mod;
      if (l > 0) {
        dp[j][l] = (dp[j - 1][l - 1] * cnt[j - 1] % mod + dp[j][l]) % mod;
      }
    }
  }
  LL ans = dp[m][k];
  for (int j = 0; j <= m; ++j) {
    for (int l = 0; l <= k; ++l) {
      dp[j][l] = 0;
    }
  }
  fill_n(cnt.begin(), m, 0);
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d%d", &n, &k);
    for (int i = 1; i < n; ++i) {
      scanf("%d%d", &u, &v);
      graph[u].push_back(v);
      graph[v].push_back(u);
    }
    if (k > 2) {
      LL ans = 0;
      for (int i = 1; i <= n; ++i) {
        fill_n(vis + 1, n, false);
        vis[i] = true;
        vector<int> cnt(graph[i].size(), 0);
        queue<tuple<int, int, int>> quu;
        for (int j = 0; j < graph[i].size(); ++j) {
          quu.push({graph[i][j], j, 1});
        }
        int nowHeight = 1;
        while (!quu.empty()) {
          auto cur = quu.front();
          quu.pop();
          vis[get<0>(cur)] = true;
          if (get<2>(cur) > nowHeight) {
            ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
            ++nowHeight;
          }
          ++cnt[get<1>(cur)];
          for (auto nxt : graph[get<0>(cur)]) {
            if (vis[nxt]) continue;
            quu.push({nxt, get<1>(cur), get<2>(cur) + 1});
          }
        }
        ans = (ans + solve(graph[i].size(), k, cnt)) % mod;
      }
      printf("%lld\n", ans);
    } else {
      printf("%lld\n", 1ll * n * (n - 1) / 2 % mod);
    }
    for (int i = 1; i <= n; ++i) {
      graph[i].clear();
    }
  }
  return 0;
}

```