#include <bits/stdc++.h>
using namespace std;
const int N = 50010, M = N + N;
const int K = 505;
int n, m;
int en, hd[N], vo[M], pv[M];
void init() {
  en = 0;
  memset(hd, -1, sizeof hd);
}
void add(int u, int v) {
  vo[en] = v;
  pv[en] = hd[u];
  hd[u] = en++;
}
int f[N][K];
long long ans;
void dfs(int u, int p) {
  f[u][0] = 1;
  for (int z = hd[u]; z > -1; z = pv[z]) {
    int v = vo[z];
    if (v == p) continue;
    dfs(v, u);
    for (int k = 1; k <= m; k++) f[u][k] += f[v][k - 1];
  }
  for (int z = hd[u]; z > -1; z = pv[z]) {
    int v = vo[z];
    if (v == p) continue;
    for (int k = 1; k <= m; k++) {
      ans += f[v][k - 1] *
             (f[u][m - k] - (m - k - 1 >= 0 ? f[v][m - k - 1] : 0)) *
             (k == m ? 2 : 1);
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  init();
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    add(u, v);
    add(v, u);
  }
  memset(f, 0, sizeof f);
  ans = 0;
  dfs(0, -1);
  printf("%I64d\n", ans / 2);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 = 50010, M = N + N;
const int K = 505;
int n, m;
int en, hd[N], vo[M], pv[M];
void init() {
  en = 0;
  memset(hd, -1, sizeof hd);
}
void add(int u, int v) {
  vo[en] = v;
  pv[en] = hd[u];
  hd[u] = en++;
}
int f[N][K];
long long ans;
void dfs(int u, int p) {
  f[u][0] = 1;
  for (int z = hd[u]; z > -1; z = pv[z]) {
    int v = vo[z];
    if (v == p) continue;
    dfs(v, u);
    for (int k = 1; k <= m; k++) f[u][k] += f[v][k - 1];
  }
  for (int z = hd[u]; z > -1; z = pv[z]) {
    int v = vo[z];
    if (v == p) continue;
    for (int k = 1; k <= m; k++) {
      ans += f[v][k - 1] *
             (f[u][m - k] - (m - k - 1 >= 0 ? f[v][m - k - 1] : 0)) *
             (k == m ? 2 : 1);
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  init();
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    add(u, v);
    add(v, u);
  }
  memset(f, 0, sizeof f);
  ans = 0;
  dfs(0, -1);
  printf("%I64d\n", ans / 2);
  return 0;
}

```