A and B and Lecture Rooms(LCA)

题目描述

A and B are preparing themselves for programming contests.

The University where A and B study is a set of rooms connected by corridors. Overall, the University has n rooms connected by n - 1 corridors so that you can get from any room to any other one by moving along the corridors. The rooms are numbered from 1 to n.

Every day А and B write contests in some rooms of their university, and after each contest they gather together in the same room and discuss problems. A and B want the distance from the rooms where problems are discussed to the rooms where contests are written to be equal. The distance between two rooms is the number of edges on the shortest path between them.

As they write contests in new rooms every day, they asked you to help them find the number of possible rooms to discuss problems for each of the following m days.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of rooms in the University.

The next n - 1 lines describe the corridors. The i-th of these lines (1 ≤ i ≤ n - 1) contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n), showing that the i-th corridor connects rooms a_i and b_i.

The next line contains integer m (1 ≤ m ≤ 10⁵) — the number of queries.

Next m lines describe the queries. The j-th of these lines (1 ≤ j ≤ m) contains two integers x_j and y_j (1 ≤ x_j, y_j ≤ n) that means that on the j-th day A will write the contest in the room x_j, B will write in the room y_j.

Output

In the i-th (1 ≤ i ≤ m) line print the number of rooms that are equidistant from the rooms where A and B write contest on the i-th day.

Examples

Input

4
1 2
1 3
2 4
1
2 3

Output

1

Input

4
1 2
2 3
2 4
2
1 2
1 3

Output

0
2
解析：
给出树上两点u,v，找到与u,v距离相等的点的个数
先找到u,v的公共祖先r,然后知道u,v间的距离，中间位置为mid；
当然若dis(u,v)为奇数则无解；
将mid的位置进行讨论：
mid在r上：
mid不在r上：

//毒瘤题 
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
vector<int>G[140000];
int n,m;
int size[140000],grand[140000][21],deep[140000];
//数据范围、数组范围要注意 
void build(int x,int pre)
{
    for(int i=1;i<=20;i++)
    {
        grand[x][i]=grand[grand[x][i-1]][i-1];
    }
    size[x]=1;
    for(int i=0;i<G[x].size();i++)
    {
        int v=G[x][i];
        if(v==pre)continue;
        deep[v]=deep[x]+1;
        grand[v][0]=x;
        build(v,x);
        size[x]+=size[v];
    }
}
int lca(int u,int v)
{
    if(deep[u]<deep[v])swap(u,v);
    int dis=deep[u]-deep[v];
/////////////////////////////////// 
    for(int k=0;k<19;k++){
        if((dis>>k)&1){    
            u=grand[u][k];
        }
    }
    //把u调到和v同一深度 
/////////////////////////////////// 
    if(u==v)return u;//当且只当u、v处于一条链时 
    for(int k=19;k>=0;k--){
        if(grand[u][k]!=grand[v][k]){
            u=grand[u][k];
            v=grand[v][k];
        }
    }
    return grand[u][0];
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);//vector可以...节约代码量 
    }
    build(1,0);//建树 
    scanf("%d",&m);//m个询问 
    for(int i=1;i<=m;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        int r=lca(u,v);
        int dis=deep[u]+deep[v]-2*deep[r];
        if(dis&1){printf("0
");}
        if(deep[u]>deep[v])swap(u,v);
        else
        {
            dis/=2;
/////////////////////////////////////////////////
            int mid=v;
            for(int k=19;k>=0;k--){
                if((dis>>k)&1) mid=grand[mid][k];
            }
/////////////////////////找到 mid 
            int ans=1;
            if(mid==r)
            {
                int preu=u,prev=v;
                int du=deep[u]-deep[r];du--;
                int dv=deep[v]-deep[r];dv--;
                for(int k=19;k>=0;k--){
                    if((du>>k)&1) preu=grand[preu][k];
                    if((dv>>k)&1) prev=grand[prev][k];
                }
                ans=n-size[preu]-size[prev];
            }
            else
            {
                int prev=v,preu=u;
                int dv=deep[v]-deep[mid];
                dv--;
                for(int k=19;k>=0;k--){
                    if((dv>>k)&1) prev=grand[prev][k];
                }
                ans=size[mid]-size[prev];
            }
            cout<<ans<<endl;
        }
    }
}