CodeForces161D： Distance in Tree（树分治）

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 "a_i b_i" (without the quotes) (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i), where a_i and b_i 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

题意：给定一棵树，问树上有多少个点对距离是K（K<=500）。

思路：明显的基础分治题，分别累计经过根节点的距离为K的点对。说他基础是以为既不需要排序，也不需要去重。复杂度O（NlogN*K）

（感悟：分治=若干个小分治+线性解决当前层 =NlogN。

              分治=若干个小分治+logN解决当前层 =NlogN*logN。

              分治=若干个小分治+K*线性解决当前层 =N*K*logN。

#include<bits/stdc++.h>
using namespace std;
const int maxn=100050;
int Laxt[maxn],Next[maxn],To[maxn];
int sz[maxn],son[maxn],root,cnt,N,K,ans,sn; //sn，小树的大小。 
int num[550],tnum[550],vis[maxn];
void read(int &x){
    x=0;     char c=getchar();
    while(c>'9'||c<'0') c=getchar();
    while(c>='0'&&c<='9') x=(x<<3)+(x<<1)+c-'0',c=getchar();
}
void add(int u,int v)
{
    Next[++cnt]=Laxt[u];
    Laxt[u]=cnt; To[cnt]=v;
} 
void getroot(int u,int fa)
{
    sz[u]=1; son[u]=0;
    for(int i=Laxt[u];i;i=Next[i]){
        int v=To[i];
        if(v==fa||vis[v]) continue;
        getroot(v,u); sz[u]+=sz[v];
        son[u]=max(son[u],sz[v]);
    }
    son[u]=max(sz[u],sn-son[u]);
    if(root==0||son[root]>son[u]) root=u;
}
void getdep(int u,int fa,int dis)
{
    if(K>=dis) ans+=num[K-dis],tnum[dis]++;
    if(K==dis)  return ;
    for(int i=Laxt[u];i;i=Next[i])
      if(To[i]!=fa&&!vis[To[i]]) 
        getdep(To[i],u,dis+1);
}
void dfs(int u)
{
    vis[u]=1;
    for(int i=1;i<=K;i++) num[i]=0; num[0]=1; 
    for(int i=Laxt[u];i;i=Next[i]){         //暴力部分 
        if(vis[To[i]]) continue;
        for(int j=0;j<=K;j++) tnum[j]=0;
        getdep(To[i],0,1);
        for(int j=0;j<=K;j++) num[j]+=tnum[j];
    }
    
    for(int i=Laxt[u];i;i=Next[i]){      //以大化小，分治部分 
        if(vis[To[i]]) continue;
        sn=sz[To[i]]; root=0; 
        getroot(To[i],u); dfs(root);
    } 
}
int main()
{
    int u,v; scanf("%d%d",&N,&K);
    for(int i=1;i<N;i++){
        read(u) ; read(v) ;
        add(u,v); add(v,u);
    }
    root=0; sn=N; getroot(1,0); dfs(root); 
    printf("%d\n",ans);
    return 0;
}