#include <bits/stdc++.h>
using namespace std;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
std::vector<int> edge[50003];
int visit[50002];
int n, k;
int done[50003];
int dp[50005][505];
int ans = 0;
void dfs(int u, int fa) {
  int i, j, v;
  dp[u][0] = 1;
  for (i = 0; i < edge[u].size(); i++) {
    v = edge[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (j = 1; j <= k; j++) ans += dp[u][j - 1] * dp[v][k - j];
    for (j = 1; j <= k; j++) dp[u][j] += dp[v][j - 1];
  }
}
int main() {
  int i, j, x, y;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  dfs(1, -1);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
std::vector<int> edge[50003];
int visit[50002];
int n, k;
int done[50003];
int dp[50005][505];
int ans = 0;
void dfs(int u, int fa) {
  int i, j, v;
  dp[u][0] = 1;
  for (i = 0; i < edge[u].size(); i++) {
    v = edge[u][i];
    if (v == fa) continue;
    dfs(v, u);
    for (j = 1; j <= k; j++) ans += dp[u][j - 1] * dp[v][k - j];
    for (j = 1; j <= k; j++) dp[u][j] += dp[v][j - 1];
  }
}
int main() {
  int i, j, x, y;
  scanf("%d%d", &n, &k);
  for (i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  dfs(1, -1);
  printf("%d\n", ans);
  return 0;
}

```