#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll A = 26;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll BASE = 10000;
double q;
using ll = long long;
inline ll bit(ll x, ll y) { return (x & (1ll << y)); }
vector<ll> a[101];
ll dp[101][101];
ll sum[10], Dp[10][10];
ll par[101];
ll n, k, kq;
void dfs(ll u) {
  dp[u][0] = 1;
  for (auto x : a[u])
    if (x != par[u]) {
      par[x] = u;
      dfs(x);
      for (ll i = 1; i <= n; i++)
        dp[u][i] = (dp[u][i] % mod + dp[x][i - 1] % mod) % mod;
    }
}
inline ll powmod(ll a, ll b) {
  ll res = 1ll;
  a %= mod;
  while (b) {
    if (b & 1ll) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1ll;
  }
  return res % mod;
}
ll d[100006], dk[100006];
void init() {
  d[0] = d[1] = dk[0] = dk[1] = 1ll;
  for (ll i = 0; i <= 100000; i++) {
    dk[i] = powmod(i, mod - 2);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  init();
  ll t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (ll i = 1; i <= n; i++) {
      a[i].clear();
    }
    for (ll i = 1; i <= n - 1; i++) {
      ll u, v;
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    if (k == 2) {
      cout << ((n - 1) * n) / 2 << endl;
      continue;
    }
    kq = 0;
    for (ll i = 1; i <= n; i++)
      if (a[i].size() >= k) {
        for (ll j = 1; j <= n; j++) par[j] = j;
        for (ll j = 1; j <= n; j++)
          for (ll jj = 1; jj <= n; jj++) {
            dp[j][jj] = 0;
          }
        dfs(i);
        for (ll j = 0; j <= n - 1; j++) {
          vector<vector<ll>> Dp(n + 1, vector<ll>(n + 1));
          vector<ll> sum(k + 1, 0ll);
          for (auto x : a[i]) {
            Dp[x][1] = dp[x][j] % mod;
            sum[1] += dp[x][j] % mod;
            sum[1] %= mod;
          }
          for (ll jj = 2; jj <= k; jj++)
            for (auto jjj : a[i]) {
              Dp[jjj][jj] = (((sum[jj - 1] * dk[jj - 1]) % mod -
                              Dp[jjj][jj - 1] + mod + mod) %
                             mod * dp[jjj][j] % mod) %
                            mod;
              sum[jj] += Dp[jjj][jj];
              sum[jj] %= mod;
            }
          kq = (kq + (sum[k] * dk[k]) % mod) % mod;
        }
      }
    cout << kq << endl;
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
const ll A = 26;
const ll inf = 1e18;
const ll mod = 1e9 + 7;
const ll BASE = 10000;
double q;
using ll = long long;
inline ll bit(ll x, ll y) { return (x & (1ll << y)); }
vector<ll> a[101];
ll dp[101][101];
ll sum[10], Dp[10][10];
ll par[101];
ll n, k, kq;
void dfs(ll u) {
  dp[u][0] = 1;
  for (auto x : a[u])
    if (x != par[u]) {
      par[x] = u;
      dfs(x);
      for (ll i = 1; i <= n; i++)
        dp[u][i] = (dp[u][i] % mod + dp[x][i - 1] % mod) % mod;
    }
}
inline ll powmod(ll a, ll b) {
  ll res = 1ll;
  a %= mod;
  while (b) {
    if (b & 1ll) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1ll;
  }
  return res % mod;
}
ll d[100006], dk[100006];
void init() {
  d[0] = d[1] = dk[0] = dk[1] = 1ll;
  for (ll i = 0; i <= 100000; i++) {
    dk[i] = powmod(i, mod - 2);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  init();
  ll t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (ll i = 1; i <= n; i++) {
      a[i].clear();
    }
    for (ll i = 1; i <= n - 1; i++) {
      ll u, v;
      cin >> u >> v;
      a[u].push_back(v);
      a[v].push_back(u);
    }
    if (k == 2) {
      cout << ((n - 1) * n) / 2 << endl;
      continue;
    }
    kq = 0;
    for (ll i = 1; i <= n; i++)
      if (a[i].size() >= k) {
        for (ll j = 1; j <= n; j++) par[j] = j;
        for (ll j = 1; j <= n; j++)
          for (ll jj = 1; jj <= n; jj++) {
            dp[j][jj] = 0;
          }
        dfs(i);
        for (ll j = 0; j <= n - 1; j++) {
          vector<vector<ll>> Dp(n + 1, vector<ll>(n + 1));
          vector<ll> sum(k + 1, 0ll);
          for (auto x : a[i]) {
            Dp[x][1] = dp[x][j] % mod;
            sum[1] += dp[x][j] % mod;
            sum[1] %= mod;
          }
          for (ll jj = 2; jj <= k; jj++)
            for (auto jjj : a[i]) {
              Dp[jjj][jj] = (((sum[jj - 1] * dk[jj - 1]) % mod -
                              Dp[jjj][jj - 1] + mod + mod) %
                             mod * dp[jjj][j] % mod) %
                            mod;
              sum[jj] += Dp[jjj][jj];
              sum[jj] %= mod;
            }
          kq = (kq + (sum[k] * dk[k]) % mod) % mod;
        }
      }
    cout << kq << endl;
  }
}

```