#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
inline long long read() {
  char ch = getchar();
  long long x = 0, r = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') r = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return r ? x : -x;
}
long long t, n, k, x, y, cnt, ans, c, head[101], maxd;
long long d[101], num[101][101];
struct node {
  long long v, next;
} e[201];
void add(long long u, long long v) {
  e[++cnt].v = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long x, long long fa, long long color) {
  d[x] = d[fa] + 1;
  maxd = max(maxd, d[x]);
  num[d[x]][color]++;
  for (long long i = head[x]; i; i = e[i].next) {
    long long v = e[i].v;
    if (v == fa) continue;
    dfs(v, x, color);
  }
}
long long dp[101][101];
void update(long long x) {
  d[x] = 1;
  c = 0;
  maxd = 1;
  for (long long i = head[x]; i; i = e[i].next) dfs(e[i].v, x, ++c);
  for (long long i = 1; i <= maxd; ++i) {
    dp[0][0] = 1;
    for (long long j = 1; j <= c; ++j) {
      dp[j][0] = 1;
      for (long long kk = 1; kk <= j; ++kk)
        dp[j][kk] =
            (dp[j - 1][kk] + (dp[j - 1][kk - 1] * num[i][j] % mod)) % mod;
    }
    ans = (ans + dp[c][k]) % mod;
    for (long long j = 1; j <= c; ++j)
      for (long long kk = 0; kk <= j; ++kk) dp[j][kk] = 0;
  }
  for (long long i = 1; i <= maxd; ++i)
    for (long long j = 1; j <= c; ++j) num[i][j] = 0;
}
signed main() {
  t = read();
  while (t--) {
    cnt = 0;
    ans = 0;
    memset(head, 0, sizeof(head));
    n = read();
    k = read();
    for (long long i = 1; i < n; ++i) {
      x = read();
      y = read();
      add(x, y);
      add(y, x);
    }
    if (k == 2)
      ans = n * (n - 1) / 2;
    else
      for (long long i = 1; i <= n; ++i) update(i);
    printf("%lld\n", ans);
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 = 1000000007;
inline long long read() {
  char ch = getchar();
  long long x = 0, r = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') r = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return r ? x : -x;
}
long long t, n, k, x, y, cnt, ans, c, head[101], maxd;
long long d[101], num[101][101];
struct node {
  long long v, next;
} e[201];
void add(long long u, long long v) {
  e[++cnt].v = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long x, long long fa, long long color) {
  d[x] = d[fa] + 1;
  maxd = max(maxd, d[x]);
  num[d[x]][color]++;
  for (long long i = head[x]; i; i = e[i].next) {
    long long v = e[i].v;
    if (v == fa) continue;
    dfs(v, x, color);
  }
}
long long dp[101][101];
void update(long long x) {
  d[x] = 1;
  c = 0;
  maxd = 1;
  for (long long i = head[x]; i; i = e[i].next) dfs(e[i].v, x, ++c);
  for (long long i = 1; i <= maxd; ++i) {
    dp[0][0] = 1;
    for (long long j = 1; j <= c; ++j) {
      dp[j][0] = 1;
      for (long long kk = 1; kk <= j; ++kk)
        dp[j][kk] =
            (dp[j - 1][kk] + (dp[j - 1][kk - 1] * num[i][j] % mod)) % mod;
    }
    ans = (ans + dp[c][k]) % mod;
    for (long long j = 1; j <= c; ++j)
      for (long long kk = 0; kk <= j; ++kk) dp[j][kk] = 0;
  }
  for (long long i = 1; i <= maxd; ++i)
    for (long long j = 1; j <= c; ++j) num[i][j] = 0;
}
signed main() {
  t = read();
  while (t--) {
    cnt = 0;
    ans = 0;
    memset(head, 0, sizeof(head));
    n = read();
    k = read();
    for (long long i = 1; i < n; ++i) {
      x = read();
      y = read();
      add(x, y);
      add(y, x);
    }
    if (k == 2)
      ans = n * (n - 1) / 2;
    else
      for (long long i = 1; i <= n; ++i) update(i);
    printf("%lld\n", ans);
  }
}

```