A and B and Lecture Rooms(LCA)

Posted 2021-02-01

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棰樼洰鎻忚堪

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 袗 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?鈮?n?鈮?10⁵) 鈥?the number of rooms in the University.

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

The next line contains integer m (1?鈮?m?鈮?10⁵) 鈥?the number of queries.

Next m lines describe the queries. The j-th of these lines (1?鈮?j?鈮?m) contains two integers x_j and y_j (1?鈮?x_j,?y_j?鈮?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?鈮?i?鈮?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
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//姣掔槫棰?
#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];
//鏁版嵁鑼冨洿銆佹暟缁勮寖鍥磋娉ㄦ剰 
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];
        }
    }
    //鎶妘璋冨埌鍜寁鍚屼竴娣卞害 
/////////////////////////////////// 
    if(u==v)return u;//褰撲笖鍙綋u銆乿澶勪簬涓€鏉￠摼鏃?
    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鍙互...鑺傜害浠ｇ爜閲?
    }
    build(1,0);//寤烘爲 
    scanf("%d",&m);//m涓闂?
    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];
            }
/////////////////////////鎵惧埌 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;
        }
    }
}