#include <bits/stdc++.h>
using namespace std;
const long long N = 5e4 + 5, M = 505;
long long lst[N], f[N][M];
long long n, m, cnt, ans;
struct edge {
  long long to, nxt;
} e[N << 1];
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
inline void add(long long u, long long v) {
  e[++cnt].to = v;
  e[cnt].nxt = lst[u];
  lst[u] = cnt;
}
inline void dfs(long long x, long long dad) {
  f[x][0] = 1;
  for (long long i = lst[x]; i; i = e[i].nxt) {
    long long to = e[i].to;
    if (to == dad) continue;
    dfs(to, x);
    for (long long j = 0; j < m; j++) ans += f[to][j] * f[x][m - j - 1];
    for (long long j = 0; j < m; j++) f[x][j + 1] += f[to][j];
  }
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i < n; i++) {
    long long u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  return printf("%lld", ans), 0;
}

### Prompt

Please formulate 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;
const long long N = 5e4 + 5, M = 505;
long long lst[N], f[N][M];
long long n, m, cnt, ans;
struct edge {
  long long to, nxt;
} e[N << 1];
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return x * f;
}
inline void add(long long u, long long v) {
  e[++cnt].to = v;
  e[cnt].nxt = lst[u];
  lst[u] = cnt;
}
inline void dfs(long long x, long long dad) {
  f[x][0] = 1;
  for (long long i = lst[x]; i; i = e[i].nxt) {
    long long to = e[i].to;
    if (to == dad) continue;
    dfs(to, x);
    for (long long j = 0; j < m; j++) ans += f[to][j] * f[x][m - j - 1];
    for (long long j = 0; j < m; j++) f[x][j + 1] += f[to][j];
  }
}
signed main() {
  n = read();
  m = read();
  for (long long i = 1; i < n; i++) {
    long long u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  return printf("%lld", ans), 0;
}

```