#include <bits/stdc++.h>
using namespace std;
int n, m, k, num, h[50010], len[50010], son[50010], tmp[50010 << 1], *id = tmp,
                                                                     *f[50010];
long long ans = 0;
struct node {
  int to, next;
} mp[50010 << 1];
inline void ins(int x, int y) {
  mp[++num].to = y;
  mp[num].next = h[x];
  h[x] = num;
  mp[++num].to = x;
  mp[num].next = h[y];
  h[y] = num;
}
void dfs1(int x, int fa) {
  for (int i = h[x]; i; i = mp[i].next) {
    int y = mp[i].to;
    if (y == fa) continue;
    dfs1(y, x);
    if (len[y] > len[son[x]]) son[x] = y;
  }
  len[x] = len[son[x]] + 1;
}
void dfs2(int x, int fa) {
  f[x][0] = 1;
  if (son[x]) {
    f[son[x]] = f[x] + 1;
    dfs2(son[x], x);
  }
  for (int i = h[x]; i; i = mp[i].next) {
    int y = mp[i].to;
    if (y == fa || y == son[x]) continue;
    f[y] = id;
    id += len[y] + 1;
    dfs2(y, x);
    for (int j = 0; j <= min(len[y], k - 2); ++j)
      if (k - j - 1 < len[x]) ans += 1ll * f[y][j] * f[x][k - j - 1];
    for (int j = 0; j <= len[y]; ++j) f[x][j + 1] += f[y][j];
  }
  if (len[x] >= k) ans += f[x][k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    ins(x, y);
  }
  dfs1(1, 0);
  f[1] = id;
  id += len[1] + 1;
  dfs2(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

### 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 n, m, k, num, h[50010], len[50010], son[50010], tmp[50010 << 1], *id = tmp,
                                                                     *f[50010];
long long ans = 0;
struct node {
  int to, next;
} mp[50010 << 1];
inline void ins(int x, int y) {
  mp[++num].to = y;
  mp[num].next = h[x];
  h[x] = num;
  mp[++num].to = x;
  mp[num].next = h[y];
  h[y] = num;
}
void dfs1(int x, int fa) {
  for (int i = h[x]; i; i = mp[i].next) {
    int y = mp[i].to;
    if (y == fa) continue;
    dfs1(y, x);
    if (len[y] > len[son[x]]) son[x] = y;
  }
  len[x] = len[son[x]] + 1;
}
void dfs2(int x, int fa) {
  f[x][0] = 1;
  if (son[x]) {
    f[son[x]] = f[x] + 1;
    dfs2(son[x], x);
  }
  for (int i = h[x]; i; i = mp[i].next) {
    int y = mp[i].to;
    if (y == fa || y == son[x]) continue;
    f[y] = id;
    id += len[y] + 1;
    dfs2(y, x);
    for (int j = 0; j <= min(len[y], k - 2); ++j)
      if (k - j - 1 < len[x]) ans += 1ll * f[y][j] * f[x][k - j - 1];
    for (int j = 0; j <= len[y]; ++j) f[x][j + 1] += f[y][j];
  }
  if (len[x] >= k) ans += f[x][k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    ins(x, y);
  }
  dfs1(1, 0);
  f[1] = id;
  id += len[1] + 1;
  dfs2(1, 0);
  printf("%I64d\n", ans);
  return 0;
}

```