#include <bits/stdc++.h>
using namespace std;
const int N = 2 * 50000 + 10;
const int KK = 500 + 10;
int n, K, i, k, j, x[N], y[N], de[N][2], dv[N][KK], u[N];
long long res, tmp;
int main() {
  scanf("%d%d", &n, &K);
  for (j = 1; j < n; j++) {
    scanf("%d%d", &x[j], &y[j]);
    x[n - 1 + j] = y[j];
    y[n - 1 + j] = x[j];
    u[n - 1 + j] = j;
    u[j] = n - 1 + j;
  }
  for (j = 1; j < 2 * n; j++) {
    de[j][1] = 1;
    dv[x[j]][1]++;
  }
  for (k = 1; k < K; k++)
    for (j = 1; j < 2 * n; j++) {
      de[j][(k + 1) % 2] = dv[y[j]][k] - de[u[j]][k % 2];
      dv[x[j]][k + 1] += de[j][(k + 1) % 2];
    }
  for (i = 1; i <= n; i++) {
    tmp = dv[i][K];
    res += tmp;
  }
  cout << res / 2LL << endl;
}

### Prompt

Please provide a cpp coded solution 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;
const int N = 2 * 50000 + 10;
const int KK = 500 + 10;
int n, K, i, k, j, x[N], y[N], de[N][2], dv[N][KK], u[N];
long long res, tmp;
int main() {
  scanf("%d%d", &n, &K);
  for (j = 1; j < n; j++) {
    scanf("%d%d", &x[j], &y[j]);
    x[n - 1 + j] = y[j];
    y[n - 1 + j] = x[j];
    u[n - 1 + j] = j;
    u[j] = n - 1 + j;
  }
  for (j = 1; j < 2 * n; j++) {
    de[j][1] = 1;
    dv[x[j]][1]++;
  }
  for (k = 1; k < K; k++)
    for (j = 1; j < 2 * n; j++) {
      de[j][(k + 1) % 2] = dv[y[j]][k] - de[u[j]][k % 2];
      dv[x[j]][k + 1] += de[j][(k + 1) % 2];
    }
  for (i = 1; i <= n; i++) {
    tmp = dv[i][K];
    res += tmp;
  }
  cout << res / 2LL << endl;
}

```