#include <bits/stdc++.h>
using namespace std;
const int N = 50001;
const int MOD = 1000000007;
const int INF = 0x3fffffff;
typedef struct Edge {
  int u, v, next;
} Edge;
Edge es[N + N];
int head[N];
int ecnt;
const int K = 501;
long long dp[N][K];
int n, k;
long long ans;
void addEdge(int u, int v) {
  es[ecnt].u = u;
  es[ecnt].v = v;
  es[ecnt].next = head[u];
  head[u] = ecnt++;
  es[ecnt].u = v;
  es[ecnt].v = u;
  es[ecnt].next = head[v];
  head[v] = ecnt++;
}
void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (int i = head[u]; i != -1; i = es[i].next) {
    if (es[i].v == fa) continue;
    int v = es[i].v;
    dfs(v, u);
    for (int j = 0; j < k; ++j) ans += dp[u][j] * dp[v][k - j - 1];
    for (int j = 1; j <= k; ++j) {
      dp[u][j] += dp[v][j - 1];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  scanf("%d%d", &n, &k);
  int u, v;
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    addEdge(u, v);
  }
  dfs(1, 0);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 int N = 50001;
const int MOD = 1000000007;
const int INF = 0x3fffffff;
typedef struct Edge {
  int u, v, next;
} Edge;
Edge es[N + N];
int head[N];
int ecnt;
const int K = 501;
long long dp[N][K];
int n, k;
long long ans;
void addEdge(int u, int v) {
  es[ecnt].u = u;
  es[ecnt].v = v;
  es[ecnt].next = head[u];
  head[u] = ecnt++;
  es[ecnt].u = v;
  es[ecnt].v = u;
  es[ecnt].next = head[v];
  head[v] = ecnt++;
}
void dfs(int u, int fa) {
  dp[u][0] = 1;
  for (int i = head[u]; i != -1; i = es[i].next) {
    if (es[i].v == fa) continue;
    int v = es[i].v;
    dfs(v, u);
    for (int j = 0; j < k; ++j) ans += dp[u][j] * dp[v][k - j - 1];
    for (int j = 1; j <= k; ++j) {
      dp[u][j] += dp[v][j - 1];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  scanf("%d%d", &n, &k);
  int u, v;
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    addEdge(u, v);
  }
  dfs(1, 0);
  printf("%d\n", ans);
  return 0;
}

```