#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char i = getchar();
  long long f = 1, res = 0;
  while (i < '0' || i > '9') {
    if (i == '-') f = -1;
    i = getchar();
  }
  while (i >= '0' && i <= '9') {
    res = res * 10 + i - '0';
    i = getchar();
  }
  return res * f;
}
const int N = 5e4 + 50;
int cnt, head[N], k, n;
long long pre[N], dp[N][505];
struct zk {
  int v, next;
} ed[N << 1];
inline void adde(int u, int v) {
  ed[++cnt].v = v;
  ed[cnt].next = head[u];
  head[u] = cnt;
}
inline void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (register int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa) continue;
    dfs(v, u);
    pre[v] = u;
    for (register int j = 1; j <= k; ++j) {
      dp[u][j] += dp[v][j - 1];
    }
  }
}
int main() {
  memset(head, -1, sizeof head);
  n = read(), k = read();
  for (register int i = 0; i < n - 1; ++i) {
    int u = read(), v = read();
    adde(u, v);
    adde(v, u);
  }
  dfs(1, -1);
  long long res = 0;
  for (register int i = 1; i <= n; ++i) {
    int now = i, last = i;
    for (register int d = 1; d <= k; ++d) {
      now = pre[now];
      if (!now) break;
      dp[i][k] += dp[now][k - d] - (k == d ? 0 : dp[last][k - d - 1]);
      last = now;
    }
    res += dp[i][k];
  }
  printf("%lld", res / 2ll);
}

### Prompt

Create a solution in CPP for 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;
inline long long read() {
  char i = getchar();
  long long f = 1, res = 0;
  while (i < '0' || i > '9') {
    if (i == '-') f = -1;
    i = getchar();
  }
  while (i >= '0' && i <= '9') {
    res = res * 10 + i - '0';
    i = getchar();
  }
  return res * f;
}
const int N = 5e4 + 50;
int cnt, head[N], k, n;
long long pre[N], dp[N][505];
struct zk {
  int v, next;
} ed[N << 1];
inline void adde(int u, int v) {
  ed[++cnt].v = v;
  ed[cnt].next = head[u];
  head[u] = cnt;
}
inline void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (register int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa) continue;
    dfs(v, u);
    pre[v] = u;
    for (register int j = 1; j <= k; ++j) {
      dp[u][j] += dp[v][j - 1];
    }
  }
}
int main() {
  memset(head, -1, sizeof head);
  n = read(), k = read();
  for (register int i = 0; i < n - 1; ++i) {
    int u = read(), v = read();
    adde(u, v);
    adde(v, u);
  }
  dfs(1, -1);
  long long res = 0;
  for (register int i = 1; i <= n; ++i) {
    int now = i, last = i;
    for (register int d = 1; d <= k; ++d) {
      now = pre[now];
      if (!now) break;
      dp[i][k] += dp[now][k - d] - (k == d ? 0 : dp[last][k - d - 1]);
      last = now;
    }
    res += dp[i][k];
  }
  printf("%lld", res / 2ll);
}

```