#include <bits/stdc++.h>
using namespace std;
long long ans;
long long opt[50010][501];
vector<int> mp[50010];
int k;
void dfs(int u, long long *best, int fa) {
  long long cur[501];
  for (int i = 0; i < mp[u].size(); i++) {
    int v = mp[u][i];
    if (v == fa) continue;
    for (int j = 1; j <= k; j++) cur[j] = 0;
    cur[0] = 1;
    dfs(v, cur, u);
    for (int j = 1; j <= k; j++) opt[u][j] += cur[j - 1];
    for (int j = 1; j <= k; j++) best[j] += cur[j - 1];
  }
  for (int i = 0; i < mp[u].size(); i++) {
    int v = mp[u][i];
    if (v == fa) continue;
    for (int j = 1; j < k; j++) {
      ans += (best[j] - opt[v][j - 1]) * opt[v][k - j - 1];
    }
    ans += 2 * opt[v][k - 1];
  }
}
int main() {
  int n, u, v;
  long long dp[501];
  while (scanf("%d %d", &n, &k) != EOF) {
    for (int i = 1; i <= n; i++) mp[i].clear();
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= k; j++) opt[i][j] = 0;
      opt[i][0] = 1;
    }
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &u, &v);
      mp[u].push_back(v);
      mp[v].push_back(u);
    }
    ans = 0, dp[0] = 1;
    for (int i = 1; i <= k; i++) dp[i] = 0;
    dfs(1, dp, -1);
    printf("%I64d\n", ans / 2);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

A tree is a connected graph that doesn't contain any cycles.

The distance between two vertices of a tree is the length (in edges) of the shortest path between these vertices.

You are given a tree with n vertices and a positive number k. Find the number of distinct pairs of the vertices which have a distance of exactly k between them. Note that pairs (v, u) and (u, v) are considered to be the same pair.

Input

The first line contains two integers n and k (1 ≤ n ≤ 50000, 1 ≤ k ≤ 500) — the number of vertices and the required distance between the vertices.

Next n - 1 lines describe the edges as "ai bi" (without the quotes) (1 ≤ ai, bi ≤ n, ai ≠ bi), where ai and bi are the vertices connected by the i-th edge. All given edges are different.

Output

Print a single integer — the number of distinct pairs of the tree's vertices which have a distance of exactly k between them.

Please do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

5 2
1 2
2 3
3 4
2 5


Output

4


Input

5 3
1 2
2 3
3 4
4 5


Output

2

Note

In the first sample the pairs of vertexes at distance 2 from each other are (1, 3), (1, 5), (3, 5) and (2, 4).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long ans;
long long opt[50010][501];
vector<int> mp[50010];
int k;
void dfs(int u, long long *best, int fa) {
  long long cur[501];
  for (int i = 0; i < mp[u].size(); i++) {
    int v = mp[u][i];
    if (v == fa) continue;
    for (int j = 1; j <= k; j++) cur[j] = 0;
    cur[0] = 1;
    dfs(v, cur, u);
    for (int j = 1; j <= k; j++) opt[u][j] += cur[j - 1];
    for (int j = 1; j <= k; j++) best[j] += cur[j - 1];
  }
  for (int i = 0; i < mp[u].size(); i++) {
    int v = mp[u][i];
    if (v == fa) continue;
    for (int j = 1; j < k; j++) {
      ans += (best[j] - opt[v][j - 1]) * opt[v][k - j - 1];
    }
    ans += 2 * opt[v][k - 1];
  }
}
int main() {
  int n, u, v;
  long long dp[501];
  while (scanf("%d %d", &n, &k) != EOF) {
    for (int i = 1; i <= n; i++) mp[i].clear();
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= k; j++) opt[i][j] = 0;
      opt[i][0] = 1;
    }
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &u, &v);
      mp[u].push_back(v);
      mp[v].push_back(u);
    }
    ans = 0, dp[0] = 1;
    for (int i = 1; i <= k; i++) dp[i] = 0;
    dfs(1, dp, -1);
    printf("%I64d\n", ans / 2);
  }
  return 0;
}

```