#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, inf = 1e9 + 7;
vector<int> vec[N];
int n, k, mx, nod, now, cnt[N], siz[N];
bool vs[N];
long long int rs;
void dfs(int u, int pr) {
  nod++, siz[u] = 1;
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    dfs(v, u);
    siz[u] += siz[v];
  }
}
int centroid(int u, int pr) {
  for (auto v : vec[u]) {
    if (v == pr || siz[v] <= nod / 2 || vs[v]) continue;
    return centroid(v, u);
  }
  return u;
}
void add(int u, int pr, int dis) {
  cnt[dis]++;
  mx = max(mx, dis);
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    add(v, u, dis + 1);
  }
}
void cal(int u, int pr, int dis) {
  if (k >= dis) {
    now += cnt[k - dis];
  } else
    return;
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    cal(v, u, dis + 1);
  }
}
void decomposed(int u, int pr) {
  nod = 0;
  dfs(u, pr);
  int rt = centroid(u, pr);
  vs[rt] = true;
  mx = 0, now = 0;
  for (auto v : vec[rt]) {
    if (v == pr || vs[v]) continue;
    cal(v, rt, 1);
    add(v, rt, 1);
  }
  rs += now;
  for (int i = 1; i <= mx; i++) {
    cnt[i] = 0;
  }
  for (auto v : vec[rt]) {
    if (v == pr || vs[v]) continue;
    decomposed(v, rt);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int m = n - 1;
  cnt[0] = 1;
  while (m--) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  decomposed(1, -1);
  cout << rs << "\n";
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const int N = 1e6 + 5, inf = 1e9 + 7;
vector<int> vec[N];
int n, k, mx, nod, now, cnt[N], siz[N];
bool vs[N];
long long int rs;
void dfs(int u, int pr) {
  nod++, siz[u] = 1;
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    dfs(v, u);
    siz[u] += siz[v];
  }
}
int centroid(int u, int pr) {
  for (auto v : vec[u]) {
    if (v == pr || siz[v] <= nod / 2 || vs[v]) continue;
    return centroid(v, u);
  }
  return u;
}
void add(int u, int pr, int dis) {
  cnt[dis]++;
  mx = max(mx, dis);
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    add(v, u, dis + 1);
  }
}
void cal(int u, int pr, int dis) {
  if (k >= dis) {
    now += cnt[k - dis];
  } else
    return;
  for (auto v : vec[u]) {
    if (v == pr || vs[v]) continue;
    cal(v, u, dis + 1);
  }
}
void decomposed(int u, int pr) {
  nod = 0;
  dfs(u, pr);
  int rt = centroid(u, pr);
  vs[rt] = true;
  mx = 0, now = 0;
  for (auto v : vec[rt]) {
    if (v == pr || vs[v]) continue;
    cal(v, rt, 1);
    add(v, rt, 1);
  }
  rs += now;
  for (int i = 1; i <= mx; i++) {
    cnt[i] = 0;
  }
  for (auto v : vec[rt]) {
    if (v == pr || vs[v]) continue;
    decomposed(v, rt);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int m = n - 1;
  cnt[0] = 1;
  while (m--) {
    int u, v;
    cin >> u >> v;
    vec[u].push_back(v);
    vec[v].push_back(u);
  }
  decomposed(1, -1);
  cout << rs << "\n";
  return 0;
}

```