#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("avx2")
using namespace std;
using pii = pair<long long, long long>;
const long long N = 1e6 + 5;
const long long INF = 1e18 + 7;
const long long MOD = 1e9 + 7;
const long long eps = 1e-6;
const long long B = 800;
vector<long long> g[N];
long long centr[N], sz[N], cnt[N], n, k, ans, sz1, mx;
void calc_sz(long long v, long long p = -1) {
  sz[v] = 1;
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      calc_sz(to, v);
      sz[v] += sz[to];
    }
  }
}
long long findc(long long v, long long p = -1) {
  for (auto to : g[v]) {
    if (to != p && !centr[to] && sz[to] >= sz1 / 2) {
      return findc(to, v);
    }
  }
  return v;
}
void calc_ans(long long v, long long p = -1, long long d = 1) {
  if (d > k) return;
  ans += cnt[k - d];
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      calc_ans(to, v, d + 1);
    }
  }
}
void add(long long v, long long p = -1, long long d = 1) {
  if (d > k) return;
  cnt[d]++;
  mx = max(mx, d);
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      add(to, v, d + 1);
    }
  }
}
void dfs(long long v) {
  calc_sz(v);
  sz1 = sz[v];
  v = findc(v);
  centr[v] = cnt[0] = 1;
  mx = 0;
  for (auto to : g[v]) {
    if (!centr[to]) {
      calc_ans(to);
      add(to);
    }
  }
  for (long long i = 0; i <= mx; i++) cnt[i] = 0;
  for (auto to : g[v]) {
    if (!centr[to]) {
      dfs(to);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  cout << ans;
  cout << '\n';
}
long long test = 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  while (test--) solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("avx2")
using namespace std;
using pii = pair<long long, long long>;
const long long N = 1e6 + 5;
const long long INF = 1e18 + 7;
const long long MOD = 1e9 + 7;
const long long eps = 1e-6;
const long long B = 800;
vector<long long> g[N];
long long centr[N], sz[N], cnt[N], n, k, ans, sz1, mx;
void calc_sz(long long v, long long p = -1) {
  sz[v] = 1;
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      calc_sz(to, v);
      sz[v] += sz[to];
    }
  }
}
long long findc(long long v, long long p = -1) {
  for (auto to : g[v]) {
    if (to != p && !centr[to] && sz[to] >= sz1 / 2) {
      return findc(to, v);
    }
  }
  return v;
}
void calc_ans(long long v, long long p = -1, long long d = 1) {
  if (d > k) return;
  ans += cnt[k - d];
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      calc_ans(to, v, d + 1);
    }
  }
}
void add(long long v, long long p = -1, long long d = 1) {
  if (d > k) return;
  cnt[d]++;
  mx = max(mx, d);
  for (auto to : g[v]) {
    if (to != p && !centr[to]) {
      add(to, v, d + 1);
    }
  }
}
void dfs(long long v) {
  calc_sz(v);
  sz1 = sz[v];
  v = findc(v);
  centr[v] = cnt[0] = 1;
  mx = 0;
  for (auto to : g[v]) {
    if (!centr[to]) {
      calc_ans(to);
      add(to);
    }
  }
  for (long long i = 0; i <= mx; i++) cnt[i] = 0;
  for (auto to : g[v]) {
    if (!centr[to]) {
      dfs(to);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(0);
  cout << ans;
  cout << '\n';
}
long long test = 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  while (test--) solve();
  return 0;
}

```