#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
const long long int mod = 1e9 + 7;
int head[maxn], sons[maxn][maxn];
long long int c[maxn][maxn], dp[maxn][maxn];
vector<int> son;
struct edge {
  int to, next;
} e[maxn * 2];
int cnt, maxdep, mindep, k;
void add(int x, int y) {
  e[cnt].to = y;
  e[cnt].next = head[x];
  head[x] = cnt++;
}
void dfs(int rt, int u, int fa, int deep) {
  sons[rt][deep]++;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(rt, v, u, deep + 1);
  }
}
long long int cal(vector<int> son) {
  memset(dp, 0, sizeof(dp));
  int n = son.size() - 1;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (j) {
        dp[i][j] += dp[i - 1][j - 1] * son[i];
      }
      dp[i][j] += dp[i - 1][j];
      dp[i][j] %= mod;
    }
  }
  return dp[n][k];
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  int t, n, x, y;
  cin >> t;
  while (t--) {
    long long int ans = 0;
    cnt = 0;
    memset(head, -1, sizeof(head));
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
      cin >> x >> y;
      add(x, y);
      add(y, x);
    }
    if (k == 2) {
      cout << c[n][2] << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      memset(sons, 0, sizeof(sons));
      for (int u = head[i]; u != -1; u = e[u].next) {
        int v = e[u].to;
        dfs(v, v, i, 1);
      }
      for (int j = 1; j <= 100; j++) {
        long long int tmp = 1;
        int tot = 0;
        son.clear();
        son.push_back(-1);
        for (int u = head[i]; u != -1; u = e[u].next) {
          int v = e[u].to;
          son.push_back(sons[v][j]);
        }
        ans = (ans + cal(son)) % mod;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const int maxn = 110;
const long long int mod = 1e9 + 7;
int head[maxn], sons[maxn][maxn];
long long int c[maxn][maxn], dp[maxn][maxn];
vector<int> son;
struct edge {
  int to, next;
} e[maxn * 2];
int cnt, maxdep, mindep, k;
void add(int x, int y) {
  e[cnt].to = y;
  e[cnt].next = head[x];
  head[x] = cnt++;
}
void dfs(int rt, int u, int fa, int deep) {
  sons[rt][deep]++;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (v == fa) continue;
    dfs(rt, v, u, deep + 1);
  }
}
long long int cal(vector<int> son) {
  memset(dp, 0, sizeof(dp));
  int n = son.size() - 1;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= i; j++) {
      if (j) {
        dp[i][j] += dp[i - 1][j - 1] * son[i];
      }
      dp[i][j] += dp[i - 1][j];
      dp[i][j] %= mod;
    }
  }
  return dp[n][k];
}
int main() {
  c[0][0] = 1;
  for (int i = 1; i < maxn; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  }
  int t, n, x, y;
  cin >> t;
  while (t--) {
    long long int ans = 0;
    cnt = 0;
    memset(head, -1, sizeof(head));
    cin >> n >> k;
    for (int i = 1; i < n; i++) {
      cin >> x >> y;
      add(x, y);
      add(y, x);
    }
    if (k == 2) {
      cout << c[n][2] << endl;
      continue;
    }
    for (int i = 1; i <= n; i++) {
      memset(sons, 0, sizeof(sons));
      for (int u = head[i]; u != -1; u = e[u].next) {
        int v = e[u].to;
        dfs(v, v, i, 1);
      }
      for (int j = 1; j <= 100; j++) {
        long long int tmp = 1;
        int tot = 0;
        son.clear();
        son.push_back(-1);
        for (int u = head[i]; u != -1; u = e[u].next) {
          int v = e[u].to;
          son.push_back(sons[v][j]);
        }
        ans = (ans + cal(son)) % mod;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```