#include <bits/stdc++.h>
using namespace std;
const long long maxn = 5e5 + 6, inf = 1e18, mod = 1e9 + 7;
long long n, k, ans, sz[maxn], h[maxn], ps[maxn];
vector<long long> adj[maxn], vec, V[maxn];
bool mark[maxn], mrk2[maxn];
void dfs1(long long u) {
  mark[u] = 1;
  sz[u] = 1;
  vec.push_back(u);
  for (long long j : adj[u])
    if (!(mark[j] || mrk2[j])) dfs1(j), sz[u] += sz[j];
}
long long get_cen(long long u, long long root, long long par) {
  for (long long j : adj[u]) {
    if (mrk2[j] || j == par) continue;
    if (sz[j] <= sz[root] / 2ll) continue;
    return get_cen(j, root, u);
  }
  return u;
}
void DFS(long long u, long long root, long long par) {
  if (V[root].size() < h[u])
    V[root].push_back(1);
  else
    V[root][h[u] - 1]++;
  ps[h[u]]++;
  for (long long j : adj[u]) {
    if (j == par || mrk2[j]) continue;
    h[j] = h[u] + 1;
    DFS(j, root, u);
  }
}
void find_cen(long long u, long long H) {
  dfs1(u);
  long long v = get_cen(u, u, u);
  mrk2[v] = 1;
  for (long long j : adj[v]) {
    if (mrk2[j]) continue;
    h[j] = 1;
    DFS(j, j, v);
  }
  for (long long j : adj[v]) {
    if (mrk2[j]) continue;
    for (long long i = 0; i < V[j].size(); i++) {
      if (i + 1 < k) {
        long long x = 0;
        if (V[j].size() >= k - i - 1) x = V[j][k - i - 2];
        ans += V[j][i] * (ps[k - i - 1] - x);
      } else if (i + 1 == k)
        ans += V[j][i] * 2ll;
    }
  }
  for (long long i : vec) {
    mark[i] = 0;
    V[i].clear();
    ps[h[i]] = 0;
    h[i] = 0;
  }
  vec.clear();
  for (long long j : adj[v])
    if (!mark[j] && !mrk2[j]) find_cen(j, H + 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  find_cen(1, 0);
  cout << ans / 2ll << endl;
}

### Prompt

Generate 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 maxn = 5e5 + 6, inf = 1e18, mod = 1e9 + 7;
long long n, k, ans, sz[maxn], h[maxn], ps[maxn];
vector<long long> adj[maxn], vec, V[maxn];
bool mark[maxn], mrk2[maxn];
void dfs1(long long u) {
  mark[u] = 1;
  sz[u] = 1;
  vec.push_back(u);
  for (long long j : adj[u])
    if (!(mark[j] || mrk2[j])) dfs1(j), sz[u] += sz[j];
}
long long get_cen(long long u, long long root, long long par) {
  for (long long j : adj[u]) {
    if (mrk2[j] || j == par) continue;
    if (sz[j] <= sz[root] / 2ll) continue;
    return get_cen(j, root, u);
  }
  return u;
}
void DFS(long long u, long long root, long long par) {
  if (V[root].size() < h[u])
    V[root].push_back(1);
  else
    V[root][h[u] - 1]++;
  ps[h[u]]++;
  for (long long j : adj[u]) {
    if (j == par || mrk2[j]) continue;
    h[j] = h[u] + 1;
    DFS(j, root, u);
  }
}
void find_cen(long long u, long long H) {
  dfs1(u);
  long long v = get_cen(u, u, u);
  mrk2[v] = 1;
  for (long long j : adj[v]) {
    if (mrk2[j]) continue;
    h[j] = 1;
    DFS(j, j, v);
  }
  for (long long j : adj[v]) {
    if (mrk2[j]) continue;
    for (long long i = 0; i < V[j].size(); i++) {
      if (i + 1 < k) {
        long long x = 0;
        if (V[j].size() >= k - i - 1) x = V[j][k - i - 2];
        ans += V[j][i] * (ps[k - i - 1] - x);
      } else if (i + 1 == k)
        ans += V[j][i] * 2ll;
    }
  }
  for (long long i : vec) {
    mark[i] = 0;
    V[i].clear();
    ps[h[i]] = 0;
    h[i] = 0;
  }
  vec.clear();
  for (long long j : adj[v])
    if (!mark[j] && !mrk2[j]) find_cen(j, H + 1);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[v].push_back(u);
    adj[u].push_back(v);
  }
  find_cen(1, 0);
  cout << ans / 2ll << endl;
}

```