#include <bits/stdc++.h>
const int N = 50010;
int par[N];
typedef struct node {
  int p;
  struct node *next;
} node;
node h[N];
void add(int x, int y) {
  node *tmp = (node *)malloc(sizeof(node));
  tmp->p = y;
  tmp->next = h[x].next;
  h[x].next = tmp;
}
int DP[N][501], k;
bool flag[N];
long long ans;
void DFS(int x) {
  flag[x] = true;
  for (node *tmp = h[x].next; tmp != NULL; tmp = tmp->next)
    if (!flag[tmp->p]) {
      par[tmp->p] = x;
      DFS(tmp->p);
      for (int i = 1; i <= k; i++) DP[x][i] += DP[tmp->p][i - 1];
    }
  DP[x][0] = 1;
  long long tmpans = 0;
  for (node *tmp = h[x].next; tmp != NULL; tmp = tmp->next)
    if (tmp->p != par[x]) {
      for (int i = 1; i < k; i++)
        tmpans += DP[tmp->p][i - 1] * (DP[x][k - i] - DP[tmp->p][k - i - 1]);
    }
  tmpans /= 2LL;
  tmpans += DP[x][k];
  ans += tmpans;
}
int main() {
  par[1] = 0;
  int n, x, y;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) h[i].next = NULL;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  ans = 0;
  DFS(1);
  printf("%d\n", ans);
  scanf(" ");
}

### Prompt

Your challenge is to write 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>
const int N = 50010;
int par[N];
typedef struct node {
  int p;
  struct node *next;
} node;
node h[N];
void add(int x, int y) {
  node *tmp = (node *)malloc(sizeof(node));
  tmp->p = y;
  tmp->next = h[x].next;
  h[x].next = tmp;
}
int DP[N][501], k;
bool flag[N];
long long ans;
void DFS(int x) {
  flag[x] = true;
  for (node *tmp = h[x].next; tmp != NULL; tmp = tmp->next)
    if (!flag[tmp->p]) {
      par[tmp->p] = x;
      DFS(tmp->p);
      for (int i = 1; i <= k; i++) DP[x][i] += DP[tmp->p][i - 1];
    }
  DP[x][0] = 1;
  long long tmpans = 0;
  for (node *tmp = h[x].next; tmp != NULL; tmp = tmp->next)
    if (tmp->p != par[x]) {
      for (int i = 1; i < k; i++)
        tmpans += DP[tmp->p][i - 1] * (DP[x][k - i] - DP[tmp->p][k - i - 1]);
    }
  tmpans /= 2LL;
  tmpans += DP[x][k];
  ans += tmpans;
}
int main() {
  par[1] = 0;
  int n, x, y;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) h[i].next = NULL;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  ans = 0;
  DFS(1);
  printf("%d\n", ans);
  scanf(" ");
}

```