#include <bits/stdc++.h>
using namespace std;
int k;
vector<vector<int>> g;
vector<int> vis, sbt, seen;
void dfs0(int u, int p) {
  sbt[u] = 1;
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfs0(v, u);
      sbt[u] += sbt[v];
    }
  }
}
int fcd(int u, int p, const int& n) {
  for (int v : g[u]) {
    if (!vis[v] && v != p && sbt[v] > n / 2) {
      return fcd(v, u, n);
    }
  }
  return u;
}
void dfsu(int u, int p, int dph = 1) {
  seen[dph]++;
  if (dph == k - 1) {
    return;
  }
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfsu(v, u, dph + 1);
    }
  }
}
long long dfsq(int u, int p, int dph = 1) {
  if (dph == k) {
    return 1;
  }
  long long r = seen[k - dph];
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      r += dfsq(v, u, dph + 1);
    }
  }
  return r;
}
void dfsc(int u, int p, int dph = 1) {
  seen[dph]--;
  if (dph == k - 1) {
    return;
  }
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfsc(v, u, dph + 1);
    }
  }
}
long long sol(int u) {
  vis[u] = 1;
  long long r = 0;
  for (int v : g[u]) {
    if (!vis[v]) {
      r += dfsq(v, u);
      dfsu(v, u);
    }
  }
  vector<int> c;
  for (int v : g[u]) {
    if (!vis[v]) {
      dfsc(v, u);
      c.push_back(fcd(v, u, sbt[v]));
    }
  }
  for (int cd : c) {
    dfs0(cd, cd);
    r += sol(cd);
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  g.resize(n);
  seen.resize(k);
  sbt.resize(n);
  vis = vector<int>(n, 0);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs0(0, 0);
  int c = fcd(0, 0, n);
  dfs0(c, c);
  cout << sol(c) << '\n';
  return 0;
}

### 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;
int k;
vector<vector<int>> g;
vector<int> vis, sbt, seen;
void dfs0(int u, int p) {
  sbt[u] = 1;
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfs0(v, u);
      sbt[u] += sbt[v];
    }
  }
}
int fcd(int u, int p, const int& n) {
  for (int v : g[u]) {
    if (!vis[v] && v != p && sbt[v] > n / 2) {
      return fcd(v, u, n);
    }
  }
  return u;
}
void dfsu(int u, int p, int dph = 1) {
  seen[dph]++;
  if (dph == k - 1) {
    return;
  }
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfsu(v, u, dph + 1);
    }
  }
}
long long dfsq(int u, int p, int dph = 1) {
  if (dph == k) {
    return 1;
  }
  long long r = seen[k - dph];
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      r += dfsq(v, u, dph + 1);
    }
  }
  return r;
}
void dfsc(int u, int p, int dph = 1) {
  seen[dph]--;
  if (dph == k - 1) {
    return;
  }
  for (int v : g[u]) {
    if (!vis[v] && v != p) {
      dfsc(v, u, dph + 1);
    }
  }
}
long long sol(int u) {
  vis[u] = 1;
  long long r = 0;
  for (int v : g[u]) {
    if (!vis[v]) {
      r += dfsq(v, u);
      dfsu(v, u);
    }
  }
  vector<int> c;
  for (int v : g[u]) {
    if (!vis[v]) {
      dfsc(v, u);
      c.push_back(fcd(v, u, sbt[v]));
    }
  }
  for (int cd : c) {
    dfs0(cd, cd);
    r += sol(cd);
  }
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n >> k;
  g.resize(n);
  seen.resize(k);
  sbt.resize(n);
  vis = vector<int>(n, 0);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs0(0, 0);
  int c = fcd(0, 0, n);
  dfs0(c, c);
  cout << sol(c) << '\n';
  return 0;
}

```