#include <bits/stdc++.h>
using namespace std;
int n, k;
struct node {
  int to, nt;
} g[50005 * 3];
int tot;
int head[50005];
void addedg(int x, int y) {
  g[tot].to = y;
  g[tot].nt = head[x];
  head[x] = tot++;
}
int sz[50005], root, Siz, f[50005];
int vis[50005];
long long ans;
long long dep[505];
void get_root(int u, int fa) {
  sz[u] = 1;
  f[u] = 0;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (v == fa || vis[v]) continue;
    get_root(v, u);
    sz[u] += sz[v];
    f[u] = max(f[u], sz[v]);
  }
  f[u] = max(f[u], Siz - f[u]);
  if (f[root] > f[u]) root = u;
}
void rdfs(int u, int fa, int de) {
  if (de > k) return;
  dep[de]++;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (v == fa || vis[v]) continue;
    rdfs(v, u, de + 1);
  }
}
long long cal(int u, int w) {
  memset(dep, 0, sizeof(dep));
  rdfs(u, 0, w);
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    if (i > k - i) break;
    if (i != k - i)
      res += dep[i] * dep[k - i];
    else
      res += dep[i] * (dep[i] - 1) / 2;
  }
  return res;
}
void dfs(int u) {
  ans += cal(u, 0);
  vis[u] = 1;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (vis[v]) continue;
    ans -= cal(v, 1);
    root = 0;
    Siz = sz[v];
    get_root(v, 0);
    dfs(root);
  }
}
int main() {
  cin >> n >> k;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d", &x), scanf("%d", &y);
    addedg(x, y);
    addedg(y, x);
  }
  f[0] = 0x3f3f3f3f;
  root = 0;
  Siz = n;
  get_root(1, 0);
  dfs(root);
  cout << ans << endl;
}

### 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;
int n, k;
struct node {
  int to, nt;
} g[50005 * 3];
int tot;
int head[50005];
void addedg(int x, int y) {
  g[tot].to = y;
  g[tot].nt = head[x];
  head[x] = tot++;
}
int sz[50005], root, Siz, f[50005];
int vis[50005];
long long ans;
long long dep[505];
void get_root(int u, int fa) {
  sz[u] = 1;
  f[u] = 0;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (v == fa || vis[v]) continue;
    get_root(v, u);
    sz[u] += sz[v];
    f[u] = max(f[u], sz[v]);
  }
  f[u] = max(f[u], Siz - f[u]);
  if (f[root] > f[u]) root = u;
}
void rdfs(int u, int fa, int de) {
  if (de > k) return;
  dep[de]++;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (v == fa || vis[v]) continue;
    rdfs(v, u, de + 1);
  }
}
long long cal(int u, int w) {
  memset(dep, 0, sizeof(dep));
  rdfs(u, 0, w);
  long long res = 0;
  for (int i = 0; i <= k; i++) {
    if (i > k - i) break;
    if (i != k - i)
      res += dep[i] * dep[k - i];
    else
      res += dep[i] * (dep[i] - 1) / 2;
  }
  return res;
}
void dfs(int u) {
  ans += cal(u, 0);
  vis[u] = 1;
  for (int i = head[u]; i + 1; i = g[i].nt) {
    int v = g[i].to;
    if (vis[v]) continue;
    ans -= cal(v, 1);
    root = 0;
    Siz = sz[v];
    get_root(v, 0);
    dfs(root);
  }
}
int main() {
  cin >> n >> k;
  memset(head, -1, sizeof(head));
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d", &x), scanf("%d", &y);
    addedg(x, y);
    addedg(y, x);
  }
  f[0] = 0x3f3f3f3f;
  root = 0;
  Siz = n;
  get_root(1, 0);
  dfs(root);
  cout << ans << endl;
}

```