#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return ((a - b) % mod + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
long long power(long long a, long long b) {
  if (b == 0) return 1LL;
  long long tmp = power(a, b / 2);
  tmp = mul(tmp, tmp);
  if (b & 1) tmp = mul(tmp, a);
  return tmp;
}
const int maxn = 105;
long long dp[maxn][maxn], fact[maxn], ifact[maxn], ans;
vector<int> adjlist[maxn];
int n, k, depth[maxn], groupsize[maxn][maxn];
void precalc() {
  fact[0] = ifact[0] = 1;
  for (int i = 1; i < maxn; i++) {
    fact[i] = mul(i, fact[i - 1]);
    ifact[i] = power(fact[i], mod - 2);
  }
}
long long ncr(long long n, long long r) {
  if (r > n) return 0LL;
  return mul(fact[n], mul(ifact[n - r], ifact[r]));
}
void pivot(int p) {
  if (adjlist[p].size() < k) return;
  memset(depth, 0, sizeof(depth));
  memset(dp, 0, sizeof(dp));
  memset(groupsize, 0, sizeof(groupsize));
  depth[p] = 1;
  queue<pair<int, int>> q;
  for (auto u : adjlist[p]) {
    q.push({u, u});
    depth[u] = 2;
  }
  while (!q.empty()) {
    pair<int, int> cur = q.front();
    q.pop();
    groupsize[cur.second][depth[cur.first]]++;
    for (auto u : adjlist[cur.first]) {
      if (depth[u] != 0) continue;
      depth[u] = depth[cur.first] + 1;
      q.push({u, cur.second});
    }
  }
  for (int i = 0; i <= adjlist[p].size(); i++) {
    dp[i][0] = 1;
  }
  for (int d = 2; d < maxn; d++) {
    int tot = 0;
    for (auto u : adjlist[p]) {
      tot += min(groupsize[u][d], 1);
    }
    if (tot < k) break;
    for (int i = 0; i < adjlist[p].size(); i++) {
      for (int j = 1; j <= min(i + 1, k); j++) {
        dp[i + 1][j] =
            add(dp[i][j], mul(dp[i][j - 1], groupsize[adjlist[p][i]][d]));
      }
    }
    ans = add(ans, dp[adjlist[p].size()][k]);
  }
}
void solve() {
  ans = 0;
  memset(dp, 0, sizeof(dp));
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    adjlist[i].clear();
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adjlist[u].push_back(v);
    adjlist[v].push_back(u);
  }
  if (k == 2) {
    cout << ncr(n, 2);
    return;
  }
  for (int i = 1; i <= n; i++) {
    pivot(i);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  precalc();
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 long long mod = 1e9 + 7;
long long add(long long a, long long b) { return (a + b) % mod; }
long long sub(long long a, long long b) { return ((a - b) % mod + mod) % mod; }
long long mul(long long a, long long b) { return (a * b) % mod; }
long long power(long long a, long long b) {
  if (b == 0) return 1LL;
  long long tmp = power(a, b / 2);
  tmp = mul(tmp, tmp);
  if (b & 1) tmp = mul(tmp, a);
  return tmp;
}
const int maxn = 105;
long long dp[maxn][maxn], fact[maxn], ifact[maxn], ans;
vector<int> adjlist[maxn];
int n, k, depth[maxn], groupsize[maxn][maxn];
void precalc() {
  fact[0] = ifact[0] = 1;
  for (int i = 1; i < maxn; i++) {
    fact[i] = mul(i, fact[i - 1]);
    ifact[i] = power(fact[i], mod - 2);
  }
}
long long ncr(long long n, long long r) {
  if (r > n) return 0LL;
  return mul(fact[n], mul(ifact[n - r], ifact[r]));
}
void pivot(int p) {
  if (adjlist[p].size() < k) return;
  memset(depth, 0, sizeof(depth));
  memset(dp, 0, sizeof(dp));
  memset(groupsize, 0, sizeof(groupsize));
  depth[p] = 1;
  queue<pair<int, int>> q;
  for (auto u : adjlist[p]) {
    q.push({u, u});
    depth[u] = 2;
  }
  while (!q.empty()) {
    pair<int, int> cur = q.front();
    q.pop();
    groupsize[cur.second][depth[cur.first]]++;
    for (auto u : adjlist[cur.first]) {
      if (depth[u] != 0) continue;
      depth[u] = depth[cur.first] + 1;
      q.push({u, cur.second});
    }
  }
  for (int i = 0; i <= adjlist[p].size(); i++) {
    dp[i][0] = 1;
  }
  for (int d = 2; d < maxn; d++) {
    int tot = 0;
    for (auto u : adjlist[p]) {
      tot += min(groupsize[u][d], 1);
    }
    if (tot < k) break;
    for (int i = 0; i < adjlist[p].size(); i++) {
      for (int j = 1; j <= min(i + 1, k); j++) {
        dp[i + 1][j] =
            add(dp[i][j], mul(dp[i][j - 1], groupsize[adjlist[p][i]][d]));
      }
    }
    ans = add(ans, dp[adjlist[p].size()][k]);
  }
}
void solve() {
  ans = 0;
  memset(dp, 0, sizeof(dp));
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    adjlist[i].clear();
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adjlist[u].push_back(v);
    adjlist[v].push_back(u);
  }
  if (k == 2) {
    cout << ncr(n, 2);
    return;
  }
  for (int i = 1; i <= n; i++) {
    pivot(i);
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  precalc();
  int t;
  cin >> t;
  while (t--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```