#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
void chmax(int &x, int y) {
  if (x < y) x = y;
}
void chmin(int &x, int y) {
  if (x > y) x = y;
}
int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ret;
}
int T, n, k;
vector<int> v[105];
int dis[105][105], C[105][105], siz[105], dp[105][105];
vector<int> s;
bool p[105];
void dfs(int x, int fa) {
  siz[x] = p[x];
  for (int(i) = (0); (i) < (v[x].size()); (i)++) {
    int to = v[x][i];
    if (to == fa) continue;
    dfs(to, x);
    siz[x] += siz[to];
  }
}
void solve() {
  for (int(i) = (1); (i) <= (n); (i)++) v[i].clear();
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++)
    for (int(j) = (1); (j) <= (n); (j)++) dis[i][j] = inf;
  for (int(i) = (1); (i) < (n); (i)++) {
    int t1, t2;
    scanf("%d%d", &t1, &t2);
    v[t1].push_back(t2);
    v[t2].push_back(t1);
    dis[t1][t2] = dis[t2][t1] = 1;
  }
  if (k == 2) {
    printf("%d\n", n * (n - 1) / 2);
    return;
  }
  for (int(k) = (1); (k) <= (n); (k)++)
    for (int(i) = (1); (i) <= (n); (i)++)
      for (int(j) = (1); (j) <= (n); (j)++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
  int ans = 0;
  for (int(root) = (1); (root) <= (n); (root)++) {
    for (int(d) = (1); (d) <= (n); (d)++) {
      int num = 0;
      s.clear();
      for (int(i) = (1); (i) <= (n); (i)++)
        if (dis[root][i] == d)
          p[i] = 1;
        else
          p[i] = 0;
      dfs(root, 0);
      for (int(i) = (0); (i) < (v[root].size()); (i)++) {
        int to = v[root][i];
        if (siz[to]) {
          num++;
          s.push_back(siz[to]);
        }
      }
      for (int(i) = (0); (i) <= (num); (i)++)
        for (int(j) = (0); (j) <= (k); (j)++) dp[i][j] = 0;
      dp[0][0] = 1;
      for (int(i) = (1); (i) <= (num); (i)++) {
        for (int(j) = (0); (j) <= (k); (j)++) {
          if (j) dp[i][j] += (long long)dp[i - 1][j - 1] * s[i - 1] % mod;
          dp[i][j] += dp[i - 1][j];
          if (dp[i][j] >= mod) dp[i][j] -= mod;
        }
      }
      ans += dp[num][k];
      if (ans >= mod) ans -= mod;
    }
  }
  printf("%d\n", ans);
}
int main() {
  C[0][0] = 1;
  for (int(i) = (1); (i) <= (100); (i)++)
    for (int(j) = (0); (j) <= (i); (j)++) {
      C[i][j] = C[i - 1][j];
      if (j) {
        C[i][j] += C[i - 1][j - 1];
        if (C[i][j] >= mod) C[i][j] -= mod;
      }
    }
  scanf("%d", &T);
  while (T--) solve();
  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 inf = 0x3f3f3f3f, mod = 1000000007;
const double pi = 3.1415926535897932, eps = 1e-6;
void chmax(int &x, int y) {
  if (x < y) x = y;
}
void chmin(int &x, int y) {
  if (x > y) x = y;
}
int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = (long long)ret * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ret;
}
int T, n, k;
vector<int> v[105];
int dis[105][105], C[105][105], siz[105], dp[105][105];
vector<int> s;
bool p[105];
void dfs(int x, int fa) {
  siz[x] = p[x];
  for (int(i) = (0); (i) < (v[x].size()); (i)++) {
    int to = v[x][i];
    if (to == fa) continue;
    dfs(to, x);
    siz[x] += siz[to];
  }
}
void solve() {
  for (int(i) = (1); (i) <= (n); (i)++) v[i].clear();
  scanf("%d%d", &n, &k);
  for (int(i) = (1); (i) <= (n); (i)++)
    for (int(j) = (1); (j) <= (n); (j)++) dis[i][j] = inf;
  for (int(i) = (1); (i) < (n); (i)++) {
    int t1, t2;
    scanf("%d%d", &t1, &t2);
    v[t1].push_back(t2);
    v[t2].push_back(t1);
    dis[t1][t2] = dis[t2][t1] = 1;
  }
  if (k == 2) {
    printf("%d\n", n * (n - 1) / 2);
    return;
  }
  for (int(k) = (1); (k) <= (n); (k)++)
    for (int(i) = (1); (i) <= (n); (i)++)
      for (int(j) = (1); (j) <= (n); (j)++)
        dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]);
  int ans = 0;
  for (int(root) = (1); (root) <= (n); (root)++) {
    for (int(d) = (1); (d) <= (n); (d)++) {
      int num = 0;
      s.clear();
      for (int(i) = (1); (i) <= (n); (i)++)
        if (dis[root][i] == d)
          p[i] = 1;
        else
          p[i] = 0;
      dfs(root, 0);
      for (int(i) = (0); (i) < (v[root].size()); (i)++) {
        int to = v[root][i];
        if (siz[to]) {
          num++;
          s.push_back(siz[to]);
        }
      }
      for (int(i) = (0); (i) <= (num); (i)++)
        for (int(j) = (0); (j) <= (k); (j)++) dp[i][j] = 0;
      dp[0][0] = 1;
      for (int(i) = (1); (i) <= (num); (i)++) {
        for (int(j) = (0); (j) <= (k); (j)++) {
          if (j) dp[i][j] += (long long)dp[i - 1][j - 1] * s[i - 1] % mod;
          dp[i][j] += dp[i - 1][j];
          if (dp[i][j] >= mod) dp[i][j] -= mod;
        }
      }
      ans += dp[num][k];
      if (ans >= mod) ans -= mod;
    }
  }
  printf("%d\n", ans);
}
int main() {
  C[0][0] = 1;
  for (int(i) = (1); (i) <= (100); (i)++)
    for (int(j) = (0); (j) <= (i); (j)++) {
      C[i][j] = C[i - 1][j];
      if (j) {
        C[i][j] += C[i - 1][j - 1];
        if (C[i][j] >= mod) C[i][j] -= mod;
      }
    }
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}

```