#include <bits/stdc++.h>
using namespace std;
long double pi = 3.14159265358979323846;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long int p = 1e9 + 7, n, k;
long long int dis[101][101];
void dfs(long long int node, long long int par, vector<long long int> v[]) {
  dis[node][0] = 1;
  for (auto it : v[node]) {
    if (it != par) {
      dfs(it, node, v);
      for (long long int i = 1; i < n; i++) dis[node][i] += dis[it][i - 1];
    }
  }
}
long long int cal(vector<long long int> &temp) {
  long long int size = temp.size(), ans = 0, i, j;
  if (size < k) return 0;
  long long int dp[size + 1][k + 1];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (i = 1; i <= size; i++) {
    dp[i][0] = 1;
    for (long long int j = 1; j <= k; j++) {
      dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * temp[i - 1]) % p;
    }
  }
  return dp[size][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int count;
    cin >> n >> k;
    count = n - 1;
    vector<long long int> v[n];
    while (count--) {
      long long int x, y;
      cin >> x >> y;
      x--, y--;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (k == 2) {
      cout << (n * (n - 1)) / 2 << "\n";
      continue;
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      queue<long long int> q;
      q.push(i);
      for (long long int x = 0; x < n; x++)
        for (long long int y = 0; y < n; y++) dis[x][y] = 0;
      dfs(i, -1, v);
      for (long long int x = 0; x < n; x++) {
        vector<long long int> temp;
        for (auto it : v[i]) temp.push_back(dis[it][x]);
        ans = (ans + cal(temp)) % p;
      }
    }
    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;
long double pi = 3.14159265358979323846;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long int p = 1e9 + 7, n, k;
long long int dis[101][101];
void dfs(long long int node, long long int par, vector<long long int> v[]) {
  dis[node][0] = 1;
  for (auto it : v[node]) {
    if (it != par) {
      dfs(it, node, v);
      for (long long int i = 1; i < n; i++) dis[node][i] += dis[it][i - 1];
    }
  }
}
long long int cal(vector<long long int> &temp) {
  long long int size = temp.size(), ans = 0, i, j;
  if (size < k) return 0;
  long long int dp[size + 1][k + 1];
  memset(dp, 0, sizeof(dp));
  dp[0][0] = 1;
  for (i = 1; i <= size; i++) {
    dp[i][0] = 1;
    for (long long int j = 1; j <= k; j++) {
      dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * temp[i - 1]) % p;
    }
  }
  return dp[size][k];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int count;
    cin >> n >> k;
    count = n - 1;
    vector<long long int> v[n];
    while (count--) {
      long long int x, y;
      cin >> x >> y;
      x--, y--;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    if (k == 2) {
      cout << (n * (n - 1)) / 2 << "\n";
      continue;
    }
    long long int ans = 0;
    for (long long int i = 0; i < n; i++) {
      queue<long long int> q;
      q.push(i);
      for (long long int x = 0; x < n; x++)
        for (long long int y = 0; y < n; y++) dis[x][y] = 0;
      dfs(i, -1, v);
      for (long long int x = 0; x < n; x++) {
        vector<long long int> temp;
        for (auto it : v[i]) temp.push_back(dis[it][x]);
        ans = (ans + cal(temp)) % p;
      }
    }
    cout << ans << "\n";
  }
}

```