#include <bits/stdc++.h>
using namespace std;
const int N = 105;
const int p = 1e9 + 7;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
struct edge {
  int to, nxt;
} e[N];
int head[N], cnte = 0;
void add_edge(int u, int v) {
  e[++cnte].to = v;
  e[cnte].nxt = head[u];
  head[u] = cnte;
}
void clear() {
  cnte = 0;
  memset(head, 0, sizeof(head));
}
vector<int> a[N];
int dp[N], deep[N], tot[N];
int dfs(int now, int father) {
  tot[deep[now] = deep[father] + 1]++;
  int ret = deep[now];
  for (int i = head[now]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == father) continue;
    ret = max(ret, dfs(to, now));
  }
  return ret;
}
signed main(signed Goodbye, char *Wangang[]) {
  (void)Goodbye, (void)Wangang;
  int T = read();
  while (T--) {
    int n = read(), k = read(), ans = 0;
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      add_edge(u, v);
      add_edge(v, u);
    }
    if (k == 2) {
      print(n * (n - 1) / 2), clear();
      continue;
    }
    for (int rt = 1; rt <= n; rt++) {
      deep[rt] = 0;
      for (int i = head[rt]; i; i = e[i].nxt) {
        int to = e[i].to, maxdeep = dfs(to, rt);
        for (int j = 1; j <= maxdeep; j++) {
          a[j].push_back(tot[j]);
          tot[j] = 0;
        }
      }
      for (int i = 1; i <= n; i++)
        if (!a[i].empty()) {
          memset(dp, 0, sizeof(dp));
          dp[0] = 1;
          for (int x : a[i])
            for (int j = n; j >= 1; j--)
              dp[j] = (dp[j] + 1LL * dp[j - 1] * x) % p;
          ans = (ans + dp[k]) % p;
          a[i].clear();
        }
    }
    print(ans);
    clear();
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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 = 105;
const int p = 1e9 + 7;
int read() {
  int s = 0;
  char c = getchar(), lc = '+';
  while (c < '0' || '9' < c) lc = c, c = getchar();
  while ('0' <= c && c <= '9') s = s * 10 + c - '0', c = getchar();
  return lc == '-' ? -s : s;
}
void write(int x) {
  if (x < 0) putchar('-'), x = -x;
  if (x < 10)
    putchar(x + '0');
  else
    write(x / 10), putchar(x % 10 + '0');
}
void print(int x, char c = '\n') {
  write(x);
  putchar(c);
}
struct edge {
  int to, nxt;
} e[N];
int head[N], cnte = 0;
void add_edge(int u, int v) {
  e[++cnte].to = v;
  e[cnte].nxt = head[u];
  head[u] = cnte;
}
void clear() {
  cnte = 0;
  memset(head, 0, sizeof(head));
}
vector<int> a[N];
int dp[N], deep[N], tot[N];
int dfs(int now, int father) {
  tot[deep[now] = deep[father] + 1]++;
  int ret = deep[now];
  for (int i = head[now]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == father) continue;
    ret = max(ret, dfs(to, now));
  }
  return ret;
}
signed main(signed Goodbye, char *Wangang[]) {
  (void)Goodbye, (void)Wangang;
  int T = read();
  while (T--) {
    int n = read(), k = read(), ans = 0;
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      add_edge(u, v);
      add_edge(v, u);
    }
    if (k == 2) {
      print(n * (n - 1) / 2), clear();
      continue;
    }
    for (int rt = 1; rt <= n; rt++) {
      deep[rt] = 0;
      for (int i = head[rt]; i; i = e[i].nxt) {
        int to = e[i].to, maxdeep = dfs(to, rt);
        for (int j = 1; j <= maxdeep; j++) {
          a[j].push_back(tot[j]);
          tot[j] = 0;
        }
      }
      for (int i = 1; i <= n; i++)
        if (!a[i].empty()) {
          memset(dp, 0, sizeof(dp));
          dp[0] = 1;
          for (int x : a[i])
            for (int j = n; j >= 1; j--)
              dp[j] = (dp[j] + 1LL * dp[j - 1] * x) % p;
          ans = (ans + dp[k]) % p;
          a[i].clear();
        }
    }
    print(ans);
    clear();
  }
  return 0;
}

```