#include <bits/stdc++.h>
using namespace std;
const long long maxn = 50050;
const long long bign = 55000;
long long n, m;
long long tmp[bign], judge[bign];
long long ans, head[maxn], sz[maxn], maxp[maxn];
long long dis[bign], size;
long long q[bign], targetnum;
long long rt, tot, vis[maxn];
struct edge {
  long long next, to, dis;
} e[maxn * 2];
inline void addedge(long long next, long long to, long long dis) {
  e[++size].to = to;
  e[size].dis = dis;
  e[size].next = head[next];
  head[next] = size;
}
void getzx(long long t, long long fat) {
  long long i, j;
  maxp[t] = sz[t] = 1;
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    if (j == fat || vis[j]) continue;
    getzx(j, t);
    sz[t] += sz[j];
    maxp[t] = max(maxp[t], sz[j]);
  }
  maxp[t] = max(maxp[t], tot - sz[t]);
  if (maxp[t] < maxp[rt]) rt = t;
}
inline void getdis(long long t, long long fat) {
  long long i, j, k;
  tmp[++tmp[0]] = dis[t];
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    k = e[i].dis;
    if (j == fat || vis[j]) continue;
    dis[j] = dis[t] + k;
    getdis(j, t);
  }
}
inline void calc(long long t) {
  long long i, j, k, l, p = 0;
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    k = e[i].dis;
    if (vis[j]) continue;
    dis[j] = k;
    tmp[0] = 0;
    getdis(j, t);
    for (k = tmp[0]; k; k--)
      if (targetnum >= tmp[k]) ans += judge[targetnum - tmp[k]];
    for (k = tmp[0]; k; k--) {
      q[++p] = tmp[k];
      judge[tmp[k]]++;
    }
  }
  for (k = p; k; k--) judge[k] = 0;
}
inline void solve(long long t) {
  long long i, j;
  vis[t] = judge[0] = 1;
  calc(t);
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    if (vis[j]) continue;
    tot = sz[j];
    maxp[rt = 0] = 0x3f3f3f3f;
    getzx(j, 0);
    solve(rt);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  register long long i, j;
  cin >> n >> targetnum;
  long long t1, t2;
  for (i = 1; i < n; i++) {
    cin >> t1 >> t2;
    addedge(t1, t2, 1);
    addedge(t2, t1, 1);
  }
  maxp[rt = 0] = 0x3f3f3f3f;
  tot = n;
  getzx(1, 0);
  solve(rt);
  cout << ans << endl;
}

### Prompt

Your challenge is to write 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 = 50050;
const long long bign = 55000;
long long n, m;
long long tmp[bign], judge[bign];
long long ans, head[maxn], sz[maxn], maxp[maxn];
long long dis[bign], size;
long long q[bign], targetnum;
long long rt, tot, vis[maxn];
struct edge {
  long long next, to, dis;
} e[maxn * 2];
inline void addedge(long long next, long long to, long long dis) {
  e[++size].to = to;
  e[size].dis = dis;
  e[size].next = head[next];
  head[next] = size;
}
void getzx(long long t, long long fat) {
  long long i, j;
  maxp[t] = sz[t] = 1;
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    if (j == fat || vis[j]) continue;
    getzx(j, t);
    sz[t] += sz[j];
    maxp[t] = max(maxp[t], sz[j]);
  }
  maxp[t] = max(maxp[t], tot - sz[t]);
  if (maxp[t] < maxp[rt]) rt = t;
}
inline void getdis(long long t, long long fat) {
  long long i, j, k;
  tmp[++tmp[0]] = dis[t];
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    k = e[i].dis;
    if (j == fat || vis[j]) continue;
    dis[j] = dis[t] + k;
    getdis(j, t);
  }
}
inline void calc(long long t) {
  long long i, j, k, l, p = 0;
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    k = e[i].dis;
    if (vis[j]) continue;
    dis[j] = k;
    tmp[0] = 0;
    getdis(j, t);
    for (k = tmp[0]; k; k--)
      if (targetnum >= tmp[k]) ans += judge[targetnum - tmp[k]];
    for (k = tmp[0]; k; k--) {
      q[++p] = tmp[k];
      judge[tmp[k]]++;
    }
  }
  for (k = p; k; k--) judge[k] = 0;
}
inline void solve(long long t) {
  long long i, j;
  vis[t] = judge[0] = 1;
  calc(t);
  for (i = head[t]; i; i = e[i].next) {
    j = e[i].to;
    if (vis[j]) continue;
    tot = sz[j];
    maxp[rt = 0] = 0x3f3f3f3f;
    getzx(j, 0);
    solve(rt);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  register long long i, j;
  cin >> n >> targetnum;
  long long t1, t2;
  for (i = 1; i < n; i++) {
    cin >> t1 >> t2;
    addedge(t1, t2, 1);
    addedge(t2, t1, 1);
  }
  maxp[rt = 0] = 0x3f3f3f3f;
  tot = n;
  getzx(1, 0);
  solve(rt);
  cout << ans << endl;
}

```