#include <bits/stdc++.h>
using namespace std;
const int M = 205;
const int MOD = 1e9 + 7;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 0;
  for (; c <= '9' && c >= '0'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  return f ? x : -x;
}
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
int n, m, num[M][M], f[M][M];
vector<int> a;
struct edge {
  int v, nxt;
} ed[M << 1];
int head[M], cnt_edge;
inline void add_edge(int u, int v) {
  ed[++cnt_edge] = (edge){v, head[u]}, head[u] = cnt_edge;
}
inline void dfs(int x, int fa) {
  num[x][0] = 1;
  for (int i = head[x]; i; i = ed[i].nxt) {
    int v = ed[i].v;
    if (v != fa) {
      dfs(v, x);
      for (int j = 1; j <= n; j++) num[x][j] += num[v][j - 1];
    }
  }
}
int main() {
  int T = read();
  while (T--) {
    n = read(), m = read();
    int ans = 0;
    cnt_edge = 0;
    for (int i = 0; i <= n; i++) head[i] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= n; j++) num[i][j] = 0;
    }
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      add_edge(u, v), add_edge(v, u);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    for (int i = 1; i <= n; i++) {
      dfs(i, 0);
      for (int j = 1; j <= n; j++) {
        for (int k = 1; k <= a.size(); k++) {
          for (int l = 1; l <= k; l++) f[k][l] = 0;
        }
        a.clear();
        for (int l = head[i]; l; l = ed[l].nxt)
          a.push_back(num[ed[l].v][j - 1]);
        f[0][0] = 1;
        for (int k = 1; k <= a.size(); k++) {
          f[k][0] = 1;
          for (int l = 1; l <= k; l++)
            f[k][l] = (f[k - 1][l] + f[k - 1][l - 1] * a[k - 1] % MOD) % MOD;
        }
        ans = (ans + f[a.size()][m]) % MOD;
      }
      for (int j = 1; j <= n; j++) {
        for (int l = 0; l <= n; l++) num[j][l] = 0;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 M = 205;
const int MOD = 1e9 + 7;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = 0;
  for (; c <= '9' && c >= '0'; c = getchar())
    x = (x << 1) + (x << 3) + (c ^ 48);
  return f ? x : -x;
}
inline int max(int a, int b) { return a > b ? a : b; }
inline int min(int a, int b) { return a < b ? a : b; }
int n, m, num[M][M], f[M][M];
vector<int> a;
struct edge {
  int v, nxt;
} ed[M << 1];
int head[M], cnt_edge;
inline void add_edge(int u, int v) {
  ed[++cnt_edge] = (edge){v, head[u]}, head[u] = cnt_edge;
}
inline void dfs(int x, int fa) {
  num[x][0] = 1;
  for (int i = head[x]; i; i = ed[i].nxt) {
    int v = ed[i].v;
    if (v != fa) {
      dfs(v, x);
      for (int j = 1; j <= n; j++) num[x][j] += num[v][j - 1];
    }
  }
}
int main() {
  int T = read();
  while (T--) {
    n = read(), m = read();
    int ans = 0;
    cnt_edge = 0;
    for (int i = 0; i <= n; i++) head[i] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= n; j++) num[i][j] = 0;
    }
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      add_edge(u, v), add_edge(v, u);
    }
    if (m == 2) {
      printf("%d\n", n * (n - 1) / 2);
      continue;
    }
    for (int i = 1; i <= n; i++) {
      dfs(i, 0);
      for (int j = 1; j <= n; j++) {
        for (int k = 1; k <= a.size(); k++) {
          for (int l = 1; l <= k; l++) f[k][l] = 0;
        }
        a.clear();
        for (int l = head[i]; l; l = ed[l].nxt)
          a.push_back(num[ed[l].v][j - 1]);
        f[0][0] = 1;
        for (int k = 1; k <= a.size(); k++) {
          f[k][0] = 1;
          for (int l = 1; l <= k; l++)
            f[k][l] = (f[k - 1][l] + f[k - 1][l - 1] * a[k - 1] % MOD) % MOD;
        }
        ans = (ans + f[a.size()][m]) % MOD;
      }
      for (int j = 1; j <= n; j++) {
        for (int l = 0; l <= n; l++) num[j][l] = 0;
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}

```