Codeforces Round #350 (Div. 2) C. Cinema

C. Cinema

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Moscow is hosting a major international conference, which is attended by n scientists from different countries. Each of the scientists knows exactly one language. For convenience, we enumerate all languages of the world with integers from 1 to 109.

In the evening after the conference, all n scientists decided to go to the cinema. There are m movies in the cinema they came to. Each of the movies is characterized by two distinct numbers — the index of audio language and the index of subtitles language. The scientist, who came to the movie, will be very pleased if he knows the audio language of the movie, will be almost satisfied if he knows the language of subtitles and will be not satisfied if he does not know neither one nor the other (note that the audio language and the subtitles language for each movie are always different).

Scientists decided to go together to the same movie. You have to help them choose the movie, such that the number of very pleased scientists is maximum possible. If there are several such movies, select among them one that will maximize the number of almost satisfied scientists.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 200 000) — the number of scientists.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the index of a language, which the i-th scientist knows.

The third line contains a positive integer m (1 ≤ m ≤ 200 000) — the number of movies in the cinema.

The fourth line contains m positive integers b1, b2, ..., bm (1 ≤ bj ≤ 109), where bj is the index of the audio language of the j-th movie.

The fifth line contains m positive integers c1, c2, ..., cm (1 ≤ cj ≤ 109), where cj is the index of subtitles language of the j-th movie.

It is guaranteed that audio languages and subtitles language are different for each movie, that is bj ≠ cj.

Output

Print the single integer — the index of a movie to which scientists should go. After viewing this movie the number of very pleased scientists should be maximum possible. If in the cinema there are several such movies, you need to choose among them one, after viewing which there will be the maximum possible number of almost satisfied scientists.

If there are several possible answers print any of them.

Examples

input

3
2 3 2
2
3 2
2 3

output

2

input

6
6 3 1 1 3 7
5
1 2 3 4 5
2 3 4 5 1

output

1

Note

In the first sample, scientists must go to the movie with the index 2, as in such case the 1-th and the 3-rd scientists will be very pleased and the 2-nd scientist will be almost satisfied.

In the second test case scientists can go either to the movie with the index 1 or the index 3. After viewing any of these movies exactlytwo scientists will be very pleased and all the others will be not satisfied.

题意：有很多科学家，每个科学家懂得一种语言，他们一起去看电影，电影声音和字幕是两种不同语言，只要能看懂字幕或者听懂语言，科学家就不会不高兴，问看哪个电影可以尽量满足更多科学家。对于多种情况的，任意输出一种即可。输入时声音一行，字幕一行，竖着两个为一组。

思路：用map处理一下各种语言的人数，判断一下同一种声音的哪种语言最多，然后在去判断在最多声音人数的情况下哪种字幕人最多。可能存在声音或者字幕科学家们都不懂的情况。

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#define ll __int64
using namespace std;

int n,m,t;
int x[200009],y[200009],pos[200009];
int a;
 map<int,int>mp;
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a);
           mp[a]++;
        }

        int mmax=0;
        int tmp=0;
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d",&x[i]);
             if(mp[x[i]] && mp[x[i]]>mmax )
             {
                 mmax=mp[x[i]];
                 tmp=x[i];
             }
        }
        //cout<<"tmp="<<tmp<<endl;
       // cout<<"mmax="<<mmax<<endl;

        int j=0;
        for(int i=0;i<m;i++)   //记录一下懂得声音最多的情况下科学家的标号
        {
             if(mp[x[i]] && mmax==mp[x[i]])
                pos[j++]=i;
        }

        for(int i=0;i<m;i++)
        {
            scanf("%d",&y[i]);
        }

        int ans=0;
        mmax=0;
        if(j==0)
        {
            for(int i=0;i<m;i++)
            {
                if(mp[y[i]] && mp[y[i]]>mmax)
                {
                    mmax=mp[y[i]];
                    ans=i;
                }
            }
            printf("%d\n",ans+1);
            continue;
        }
        ans=pos[0];
        mmax=mp[y[ans]];

        for(int i=0;i<j;i++)
        {
            if(mp[y[pos[i]]] && mp[y[pos[i]]]>mmax)
            {
                mmax=mp[y[pos[i]]];
                ans=pos[i];    //注意这里输出位置
            }
        }
        printf("%d\n",ans+1);
    }
    return 0;
}