#include <bits/stdc++.h>
using namespace std;
const int N = 100 + 7;
const long long MOD = 1E9 + 7;
long long res = 0, fact[N], rev[N], dp[N][N];
vector<int> G[N];
int timer = 0, cnt[N], tin[N], depth[N], sum[N];
void add(int pos, int val) {
  if (tin[pos] != timer) {
    tin[pos] = timer;
    cnt[pos] = 0;
  }
  cnt[pos] += val;
}
int get(int pos) {
  if (tin[pos] != timer) {
    tin[pos] = timer;
    cnt[pos] = 0;
  }
  return cnt[pos];
}
void dfs(int v, int par) {
  depth[v] = depth[par] + 1;
  add(depth[v], 1);
  for (int to : G[v]) {
    if (to == par) continue;
    dfs(to, v);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  res = 0;
  for (int root = 1; root <= n; ++root) {
    if (G[root].size() < k) continue;
    for (int i = 0; i <= n; ++i) {
      for (int j = 1; j <= k; ++j) dp[i][j] = 0;
      dp[i][0] = 1;
      sum[i] = 0;
    }
    depth[root] = 0;
    for (int v : G[root]) {
      ++timer;
      dfs(v, root);
      for (int dep = 1; dep <= n; ++dep) {
        if (get(dep) == 0) continue;
        for (int ct = min(k - 1, sum[dep]); ct >= 0; --ct) {
          dp[dep][ct + 1] = (dp[dep][ct + 1] + dp[dep][ct] * get(dep)) % MOD;
        }
        sum[dep] = min(sum[dep] + 1, k);
      }
    }
    for (int dep = 1; dep <= n; ++dep) res = (res + dp[dep][k]) % MOD;
  }
  if (k == 2) {
    res = long long(n) * (n - 1) / 2 % MOD;
  }
  if (k == 1) {
    res = n;
  }
  cout << res << endl;
  for (int i = 1; i <= n; ++i) G[i].clear();
}
long long binpow(long long x, int power) {
  long long mult = x % MOD;
  x = 1;
  while (power) {
    if (power & 1) x = x * mult % MOD;
    mult = mult * mult % MOD;
    power = power >> 1;
  }
  return x;
}
int main() {
  fact[0] = rev[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = fact[i - 1] * i % MOD;
    rev[i] = binpow(fact[i], MOD - 2);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Generate a CPP solution 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 N = 100 + 7;
const long long MOD = 1E9 + 7;
long long res = 0, fact[N], rev[N], dp[N][N];
vector<int> G[N];
int timer = 0, cnt[N], tin[N], depth[N], sum[N];
void add(int pos, int val) {
  if (tin[pos] != timer) {
    tin[pos] = timer;
    cnt[pos] = 0;
  }
  cnt[pos] += val;
}
int get(int pos) {
  if (tin[pos] != timer) {
    tin[pos] = timer;
    cnt[pos] = 0;
  }
  return cnt[pos];
}
void dfs(int v, int par) {
  depth[v] = depth[par] + 1;
  add(depth[v], 1);
  for (int to : G[v]) {
    if (to == par) continue;
    dfs(to, v);
  }
}
void solve() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  res = 0;
  for (int root = 1; root <= n; ++root) {
    if (G[root].size() < k) continue;
    for (int i = 0; i <= n; ++i) {
      for (int j = 1; j <= k; ++j) dp[i][j] = 0;
      dp[i][0] = 1;
      sum[i] = 0;
    }
    depth[root] = 0;
    for (int v : G[root]) {
      ++timer;
      dfs(v, root);
      for (int dep = 1; dep <= n; ++dep) {
        if (get(dep) == 0) continue;
        for (int ct = min(k - 1, sum[dep]); ct >= 0; --ct) {
          dp[dep][ct + 1] = (dp[dep][ct + 1] + dp[dep][ct] * get(dep)) % MOD;
        }
        sum[dep] = min(sum[dep] + 1, k);
      }
    }
    for (int dep = 1; dep <= n; ++dep) res = (res + dp[dep][k]) % MOD;
  }
  if (k == 2) {
    res = long long(n) * (n - 1) / 2 % MOD;
  }
  if (k == 1) {
    res = n;
  }
  cout << res << endl;
  for (int i = 1; i <= n; ++i) G[i].clear();
}
long long binpow(long long x, int power) {
  long long mult = x % MOD;
  x = 1;
  while (power) {
    if (power & 1) x = x * mult % MOD;
    mult = mult * mult % MOD;
    power = power >> 1;
  }
  return x;
}
int main() {
  fact[0] = rev[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = fact[i - 1] * i % MOD;
    rev[i] = binpow(fact[i], MOD - 2);
  }
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```