#include <bits/stdc++.h>
using namespace std;
struct node {
  map<long long, long long> st;
  node() { st = {}; }
};
const long long maxn = 100;
vector<long long> g[maxn];
long long h[maxn];
const long long mod = 1e9 + 7;
vector<node *> all;
node *dfs(long long v, long long p) {
  if (p == -1)
    h[v] = 0;
  else
    h[v] = h[p] + 1;
  vector<long long> ans;
  vector<node *> gg;
  for (long long i = 0; i < g[v].size(); ++i) {
    long long to = g[v][i];
    if (to == p) continue;
    gg.push_back(dfs(to, v));
  }
  if (h[v] != 0) {
    node *res = new node();
    long long id = 0;
    for (long long j = 0; j < gg.size(); ++j) {
      if (gg[id]->st.size() < gg[j]->st.size()) id = j;
    }
    if (gg.size() != 0) res = gg[id];
    res->st[(h[v])]++;
    for (long long k = 0; k < gg.size(); ++k) {
      if (k == id) continue;
      for (auto y : gg[k]->st) {
        res->st[(y).first] += y.second;
      }
    }
    if (h[v] == 1) all.push_back(res);
    return res;
  }
  return new node();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long q;
  cin >> q;
  while (q--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n; ++i) {
      g[i].clear();
    }
    for (long long m = 0; m < n - 1; ++m) {
      long long a, b;
      cin >> a >> b;
      a--;
      b--;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 % mod << "\n";
      continue;
    }
    long long ans = 0;
    for (long long j = 0; j < n; ++j) {
      all.clear();
      dfs(j, -1);
      for (long long i = 1; i < n; ++i) {
        vector<long long> kol;
        for (long long l = 0; l < all.size(); ++l) {
          if (all[l]->st.count(i)) kol.push_back(all[l]->st[i]);
        }
        if (kol.size() < k) continue;
        long long dp[kol.size() + 1][k + 1];
        for (long long i1 = 0; i1 < kol.size() + 1; ++i1) {
          for (long long l = 0; l < k + 1; ++l) {
            dp[i1][l] = 0;
          }
        }
        dp[0][0] = 1;
        for (long long m = 1; m < kol.size() + 1; ++m) {
          dp[m][0] = 1;
          for (long long l = 1; l < k + 1; ++l) {
            dp[m][l] += dp[m - 1][l];
            dp[m][l] += dp[m - 1][l - 1] * kol[m - 1] % mod;
            dp[m][l] %= mod;
          }
        }
        ans = (ans + dp[kol.size()][k]) % mod;
      }
    }
    cout << ans << "\n";
  }
}

### 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;
struct node {
  map<long long, long long> st;
  node() { st = {}; }
};
const long long maxn = 100;
vector<long long> g[maxn];
long long h[maxn];
const long long mod = 1e9 + 7;
vector<node *> all;
node *dfs(long long v, long long p) {
  if (p == -1)
    h[v] = 0;
  else
    h[v] = h[p] + 1;
  vector<long long> ans;
  vector<node *> gg;
  for (long long i = 0; i < g[v].size(); ++i) {
    long long to = g[v][i];
    if (to == p) continue;
    gg.push_back(dfs(to, v));
  }
  if (h[v] != 0) {
    node *res = new node();
    long long id = 0;
    for (long long j = 0; j < gg.size(); ++j) {
      if (gg[id]->st.size() < gg[j]->st.size()) id = j;
    }
    if (gg.size() != 0) res = gg[id];
    res->st[(h[v])]++;
    for (long long k = 0; k < gg.size(); ++k) {
      if (k == id) continue;
      for (auto y : gg[k]->st) {
        res->st[(y).first] += y.second;
      }
    }
    if (h[v] == 1) all.push_back(res);
    return res;
  }
  return new node();
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  long long q;
  cin >> q;
  while (q--) {
    long long n, k;
    cin >> n >> k;
    for (long long i = 0; i < n; ++i) {
      g[i].clear();
    }
    for (long long m = 0; m < n - 1; ++m) {
      long long a, b;
      cin >> a >> b;
      a--;
      b--;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 % mod << "\n";
      continue;
    }
    long long ans = 0;
    for (long long j = 0; j < n; ++j) {
      all.clear();
      dfs(j, -1);
      for (long long i = 1; i < n; ++i) {
        vector<long long> kol;
        for (long long l = 0; l < all.size(); ++l) {
          if (all[l]->st.count(i)) kol.push_back(all[l]->st[i]);
        }
        if (kol.size() < k) continue;
        long long dp[kol.size() + 1][k + 1];
        for (long long i1 = 0; i1 < kol.size() + 1; ++i1) {
          for (long long l = 0; l < k + 1; ++l) {
            dp[i1][l] = 0;
          }
        }
        dp[0][0] = 1;
        for (long long m = 1; m < kol.size() + 1; ++m) {
          dp[m][0] = 1;
          for (long long l = 1; l < k + 1; ++l) {
            dp[m][l] += dp[m - 1][l];
            dp[m][l] += dp[m - 1][l - 1] * kol[m - 1] % mod;
            dp[m][l] %= mod;
          }
        }
        ans = (ans + dp[kol.size()][k]) % mod;
      }
    }
    cout << ans << "\n";
  }
}

```