#include <bits/stdc++.h>
using namespace std;
int k;
int first[110000], nxt[210000], vv[210000];
int num[50010][510], vis[50010];
long long ans[50010];
void dfs(int n) {
  num[n][0] = 1;
  vis[n] = 1;
  for (int i = 1; i <= k; i++) num[n][i] = 0;
  ans[n] = 0;
  for (int e = first[n]; e; e = nxt[e])
    if (!vis[vv[e]]) {
      dfs(vv[e]);
      for (int i = 0; i <= k; i++) {
        num[n][i + 1] += num[vv[e]][i];
        if (i > 0 && k - i > 0)
          ans[n] -= (num[vv[e]][i - 1] + 0ll) * num[vv[e]][k - i - 1];
      }
    }
  for (int i = 0; i <= k; i++) ans[n] += (num[n][i] + 0ll) * num[n][k - i];
}
int main() {
  int n, i, j, e = 2;
  while (cin >> n >> k) {
    e = 2;
    memset(first, 0, sizeof(first));
    memset(vis, 0, sizeof(vis));
    int u, v;
    for (i = 1; i < n; i++) {
      scanf("%d%d", &u, &v);
      nxt[e] = first[u], vv[e] = v, first[u] = e++;
      nxt[e] = first[v], vv[e] = u, first[v] = e++;
    }
    dfs((n + 1) / 2);
    long long aans = 0;
    for (i = 1; i <= n; i++) aans += ans[i];
    cout << aans / 2 << endl;
  }
}

### Prompt

Generate 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;
int k;
int first[110000], nxt[210000], vv[210000];
int num[50010][510], vis[50010];
long long ans[50010];
void dfs(int n) {
  num[n][0] = 1;
  vis[n] = 1;
  for (int i = 1; i <= k; i++) num[n][i] = 0;
  ans[n] = 0;
  for (int e = first[n]; e; e = nxt[e])
    if (!vis[vv[e]]) {
      dfs(vv[e]);
      for (int i = 0; i <= k; i++) {
        num[n][i + 1] += num[vv[e]][i];
        if (i > 0 && k - i > 0)
          ans[n] -= (num[vv[e]][i - 1] + 0ll) * num[vv[e]][k - i - 1];
      }
    }
  for (int i = 0; i <= k; i++) ans[n] += (num[n][i] + 0ll) * num[n][k - i];
}
int main() {
  int n, i, j, e = 2;
  while (cin >> n >> k) {
    e = 2;
    memset(first, 0, sizeof(first));
    memset(vis, 0, sizeof(vis));
    int u, v;
    for (i = 1; i < n; i++) {
      scanf("%d%d", &u, &v);
      nxt[e] = first[u], vv[e] = v, first[u] = e++;
      nxt[e] = first[v], vv[e] = u, first[v] = e++;
    }
    dfs((n + 1) / 2);
    long long aans = 0;
    for (i = 1; i <= n; i++) aans += ans[i];
    cout << aans / 2 << endl;
  }
}

```