#include <bits/stdc++.h>
using namespace std;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int32_t MOD = 1e9 + 7;
int32_t add(int32_t a, int32_t b) {
  if (a + b < MOD) return a + b;
  return a + b - MOD;
}
int32_t sub(int32_t a, int32_t b) {
  if (a - b >= 0) return a - b;
  return a - b + MOD;
}
int32_t mul(int32_t a, int32_t b) { return int64_t(a) * b % MOD; }
vector<vector<int32_t>> g;
int32_t n;
void dfs(vector<vector<int32_t>> &cntH, int32_t v, int32_t p) {
  cntH[v][0] = 1;
  for (auto &u : g[v]) {
    if (u != p) {
      dfs(cntH, u, v);
      for (int32_t i = 1; i < n; ++i) {
        cntH[v][i] += cntH[u][i - 1];
      }
    }
  }
}
void solve() {
  int32_t k;
  cin >> n >> k;
  int32_t ans = 0;
  g.clear();
  g.resize(n);
  for (int32_t i = 0; i + 1 < n; ++i) {
    int32_t 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 << '\n';
    return;
  }
  for (int32_t v = 0; v < n; ++v) {
    vector<vector<int32_t>> cntH(n, vector<int32_t>(n, 0));
    dfs(cntH, v, v);
    int32_t m = g[v].size();
    for (int32_t h = 0; h < n; ++h) {
      vector<int32_t> cnt;
      cnt.reserve(m);
      for (auto &u : g[v]) {
        cnt.push_back(cntH[u][h]);
      }
      vector<vector<int32_t>> dp(m + 1, vector<int32_t>(k + 1, 0));
      dp[0][0] = 1;
      for (int32_t i = 0; i < m; ++i) {
        dp[i + 1][0] = 1;
        for (int32_t j = 1; j <= k; ++j) {
          dp[i + 1][j] =
              add(dp[i + 1][j], add(dp[i][j], mul(dp[i][j - 1], cnt[i])));
        }
      }
      ans = add(ans, dp[m][k]);
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  io();
  int32_t ttt;
  std::cin >> ttt;
  while (ttt-- > 0) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution 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;
void io() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
const int32_t MOD = 1e9 + 7;
int32_t add(int32_t a, int32_t b) {
  if (a + b < MOD) return a + b;
  return a + b - MOD;
}
int32_t sub(int32_t a, int32_t b) {
  if (a - b >= 0) return a - b;
  return a - b + MOD;
}
int32_t mul(int32_t a, int32_t b) { return int64_t(a) * b % MOD; }
vector<vector<int32_t>> g;
int32_t n;
void dfs(vector<vector<int32_t>> &cntH, int32_t v, int32_t p) {
  cntH[v][0] = 1;
  for (auto &u : g[v]) {
    if (u != p) {
      dfs(cntH, u, v);
      for (int32_t i = 1; i < n; ++i) {
        cntH[v][i] += cntH[u][i - 1];
      }
    }
  }
}
void solve() {
  int32_t k;
  cin >> n >> k;
  int32_t ans = 0;
  g.clear();
  g.resize(n);
  for (int32_t i = 0; i + 1 < n; ++i) {
    int32_t 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 << '\n';
    return;
  }
  for (int32_t v = 0; v < n; ++v) {
    vector<vector<int32_t>> cntH(n, vector<int32_t>(n, 0));
    dfs(cntH, v, v);
    int32_t m = g[v].size();
    for (int32_t h = 0; h < n; ++h) {
      vector<int32_t> cnt;
      cnt.reserve(m);
      for (auto &u : g[v]) {
        cnt.push_back(cntH[u][h]);
      }
      vector<vector<int32_t>> dp(m + 1, vector<int32_t>(k + 1, 0));
      dp[0][0] = 1;
      for (int32_t i = 0; i < m; ++i) {
        dp[i + 1][0] = 1;
        for (int32_t j = 1; j <= k; ++j) {
          dp[i + 1][j] =
              add(dp[i + 1][j], add(dp[i][j], mul(dp[i][j - 1], cnt[i])));
        }
      }
      ans = add(ans, dp[m][k]);
    }
  }
  cout << ans << '\n';
}
int32_t main() {
  io();
  int32_t ttt;
  std::cin >> ttt;
  while (ttt-- > 0) {
    solve();
  }
  return 0;
}

```