#include <bits/stdc++.h>
using namespace std;
vector<int> G[50005];
int n, K;
long long ans;
bool vis[50005];
int judge[50005];
int rt, siz[50005], maxn[50005], tot, dis[50005];
void findRoot(int u, int fa) {
  siz[u] = 1;
  maxn[u] = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa || vis[v]) continue;
    findRoot(v, u);
    siz[u] += siz[v];
    maxn[u] = max(maxn[u], siz[v]);
  }
  maxn[u] = max(maxn[u], tot - siz[u]);
  if (maxn[u] < maxn[rt]) rt = u;
}
int tmp[50005];
void get_dis(int u, int fa) {
  tmp[++tmp[0]] = dis[u];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa || vis[v]) continue;
    dis[v] = dis[u] + 1;
    get_dis(v, u);
  }
}
int in[50005];
void calc(int u) {
  int cnt = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v]) continue;
    dis[v] = 1;
    tmp[0] = 0;
    get_dis(v, u);
    for (int j = 1; j <= tmp[0]; j++) {
      if (K >= tmp[j]) ans += judge[K - tmp[j]];
    }
    for (int j = 1; j <= tmp[0]; j++) in[++cnt] = tmp[j], judge[tmp[j]]++;
  }
  for (int i = 1; i <= cnt; i++) judge[in[i]] = 0;
}
void solve(int u) {
  vis[u] = judge[0] = 1;
  calc(u);
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v]) continue;
    tot = siz[v];
    maxn[rt = 0] = 1e9;
    findRoot(v, 0);
    solve(rt);
  }
}
int main() {
  scanf("%d %d", &n, &K);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  maxn[rt = 0] = n;
  tot = n;
  findRoot(1, 0);
  solve(rt);
  printf("%lld", ans);
  return 0;
}

### 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;
vector<int> G[50005];
int n, K;
long long ans;
bool vis[50005];
int judge[50005];
int rt, siz[50005], maxn[50005], tot, dis[50005];
void findRoot(int u, int fa) {
  siz[u] = 1;
  maxn[u] = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa || vis[v]) continue;
    findRoot(v, u);
    siz[u] += siz[v];
    maxn[u] = max(maxn[u], siz[v]);
  }
  maxn[u] = max(maxn[u], tot - siz[u]);
  if (maxn[u] < maxn[rt]) rt = u;
}
int tmp[50005];
void get_dis(int u, int fa) {
  tmp[++tmp[0]] = dis[u];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa || vis[v]) continue;
    dis[v] = dis[u] + 1;
    get_dis(v, u);
  }
}
int in[50005];
void calc(int u) {
  int cnt = 0;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v]) continue;
    dis[v] = 1;
    tmp[0] = 0;
    get_dis(v, u);
    for (int j = 1; j <= tmp[0]; j++) {
      if (K >= tmp[j]) ans += judge[K - tmp[j]];
    }
    for (int j = 1; j <= tmp[0]; j++) in[++cnt] = tmp[j], judge[tmp[j]]++;
  }
  for (int i = 1; i <= cnt; i++) judge[in[i]] = 0;
}
void solve(int u) {
  vis[u] = judge[0] = 1;
  calc(u);
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v]) continue;
    tot = siz[v];
    maxn[rt = 0] = 1e9;
    findRoot(v, 0);
    solve(rt);
  }
}
int main() {
  scanf("%d %d", &n, &K);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  maxn[rt = 0] = n;
  tot = n;
  findRoot(1, 0);
  solve(rt);
  printf("%lld", ans);
  return 0;
}

```