#include <bits/stdc++.h>
using namespace std;
int n, k, a, b;
long long dp[50007][507], ans;
vector<int> e[50007];
void add(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void Clear() {
  for (int i = 0; i < 50007; i++) e[i].clear();
}
void dfs(int u, int p) {
  dp[u][0] = 1;
  for (int i = 1; i <= k; i++) dp[u][i] = 0;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = 1; j <= k; j++) dp[u][j] += dp[v][j - 1];
  }
}
void solve(int u, int p) {
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    for (int j = k; j >= 1; j--) {
      dp[v][j] += dp[u][j - 1];
      if (j > 1) dp[v][j] -= dp[v][j - 2];
    }
    solve(v, u);
  }
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    Clear();
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &a, &b);
      add(a, b);
    }
    ans = 0;
    dfs(1, -1);
    solve(1, -1);
    for (int i = 1; i <= n; i++) ans += dp[i][k];
    printf("%I64d\n", ans / 2LL);
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int n, k, a, b;
long long dp[50007][507], ans;
vector<int> e[50007];
void add(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void Clear() {
  for (int i = 0; i < 50007; i++) e[i].clear();
}
void dfs(int u, int p) {
  dp[u][0] = 1;
  for (int i = 1; i <= k; i++) dp[u][i] = 0;
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    dfs(v, u);
    for (int j = 1; j <= k; j++) dp[u][j] += dp[v][j - 1];
  }
}
void solve(int u, int p) {
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == p) continue;
    for (int j = k; j >= 1; j--) {
      dp[v][j] += dp[u][j - 1];
      if (j > 1) dp[v][j] -= dp[v][j - 2];
    }
    solve(v, u);
  }
}
int main() {
  while (~scanf("%d%d", &n, &k)) {
    Clear();
    for (int i = 1; i < n; i++) {
      scanf("%d%d", &a, &b);
      add(a, b);
    }
    ans = 0;
    dfs(1, -1);
    solve(1, -1);
    for (int i = 1; i <= n; i++) ans += dp[i][k];
    printf("%I64d\n", ans / 2LL);
  }
}

```