#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;
constexpr int N = 101;
constexpr int M = 1e9 + 7;
ll ans;
int n, k;
vector<int> edges[N];
int depth[N][N];
int arr[N];
int dp[N][N];
void dfs(int cur, int par) {
  memset(depth[cur], 0, sizeof(int) * n);
  depth[cur][0] = 1;
  for (int i = (0); i < (edges[cur].size()); ++i) {
    int nxt = edges[cur][i];
    if (nxt == par) continue;
    dfs(nxt, cur);
    for (int j = (1); j < (n); ++j) {
      depth[cur][j] += depth[nxt][j - 1];
    }
  }
}
void solve(int root) {
  if (edges[root].size() < k) return;
  dfs(root, -1);
  for (int dist = (0); dist < (n); ++dist) {
    for (int i = (0); i < (edges[root].size()); ++i) {
      int child = edges[root][i];
      int x = depth[child][dist];
      arr[i] = x;
    }
    memset(dp, 0, sizeof(int) * N * N);
    for (int i = (0); i < (edges[root].size()); ++i) {
      dp[i][0] = 1;
    }
    for (int i = (1); i <= (edges[root].size()); ++i) {
      for (int j = (1); j <= (min(i, k)); ++j) {
        dp[i][j] = (dp[i - 1][j] + (ll)arr[i - 1] * dp[i - 1][j - 1]) % M;
      }
    }
    ans += dp[edges[root].size()][k];
    ans %= M;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    memset(edges, 0, sizeof(vector<int>) * N);
    ans = 0;
    for (int i = (0); i < (n - 1); ++i) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      edges[u].push_back(v);
      edges[v].push_back(u);
    }
    if (k == 2) {
      ans = n * (n - 1) / 2;
      ans %= M;
    } else {
      for (int i = (0); i < (n); ++i) {
        solve(i);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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;
using ll = long long;
using pii = pair<int, int>;
constexpr int N = 101;
constexpr int M = 1e9 + 7;
ll ans;
int n, k;
vector<int> edges[N];
int depth[N][N];
int arr[N];
int dp[N][N];
void dfs(int cur, int par) {
  memset(depth[cur], 0, sizeof(int) * n);
  depth[cur][0] = 1;
  for (int i = (0); i < (edges[cur].size()); ++i) {
    int nxt = edges[cur][i];
    if (nxt == par) continue;
    dfs(nxt, cur);
    for (int j = (1); j < (n); ++j) {
      depth[cur][j] += depth[nxt][j - 1];
    }
  }
}
void solve(int root) {
  if (edges[root].size() < k) return;
  dfs(root, -1);
  for (int dist = (0); dist < (n); ++dist) {
    for (int i = (0); i < (edges[root].size()); ++i) {
      int child = edges[root][i];
      int x = depth[child][dist];
      arr[i] = x;
    }
    memset(dp, 0, sizeof(int) * N * N);
    for (int i = (0); i < (edges[root].size()); ++i) {
      dp[i][0] = 1;
    }
    for (int i = (1); i <= (edges[root].size()); ++i) {
      for (int j = (1); j <= (min(i, k)); ++j) {
        dp[i][j] = (dp[i - 1][j] + (ll)arr[i - 1] * dp[i - 1][j - 1]) % M;
      }
    }
    ans += dp[edges[root].size()][k];
    ans %= M;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    memset(edges, 0, sizeof(vector<int>) * N);
    ans = 0;
    for (int i = (0); i < (n - 1); ++i) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      edges[u].push_back(v);
      edges[v].push_back(u);
    }
    if (k == 2) {
      ans = n * (n - 1) / 2;
      ans %= M;
    } else {
      for (int i = (0); i < (n); ++i) {
        solve(i);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```