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 exactly twoscientists will be very pleased and all the others will be not satisfied.

题意：有n个科学家去看电影，每个科学都只会一种语言。他们可以选择看的电影有m个。电影的配音和字幕是不同的语言，如果一个科学能听懂配音，那他就会非常高兴。如果他能看懂字母，则他会觉得满意。如果两者都不行，他就会很失望。问你选择哪个电影，非常高兴地人很多 。如果有多种情况，输出满意的人较多的情况。如果也有多种，则输出任意一种都行。

输入：n表示科学家人数

           接下来n个数，表示科学家会的语言。

   m表示电影种数

  接下来m个数表示电影的配音

           接下来m个数表示电影的字幕

输出：输出高兴的人最多的电影

解法：我们可以用一个map映射会某种语言的人数，然后遍历每部电影，按要求不断更新人数最多的人即可。

AC:
#include <stdio.h>
#include <algorithm>
#include<iostream>
#include<map>
using namespace std;
__int64 b[200018],c[200018];
int main()
{
__int64 n,t,v,mm;
__int64 a,bb,ans;
a=bb=ans=0;
ans=1;
    map<__int64,__int64>m;
    m.clear();
    scanf("%I64d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%I64d",&v);
        if(m.find(v)==m.end())
        {
            m[v]=1;
        }
        else m[v]++;
    }
    scanf("%I64d",&mm);
     for(int i=0;i<mm;i++)
    {
        scanf("%I64d",&b[i]);
    }
     for(int i=0;i<mm;i++)
    {
        scanf("%I64d",&c[i]);
    }
    for(int i=0;i<mm;i++)
    {
       if(m[b[i]]>a)
       {
           a=m[b[i]];
           ans=i+1;
           bb=m[c[i]];
       }
       else if(m[b[i]]==a)
       {
           if(m[c[i]]>bb)
           {
               bb=m[c[i]];
               ans=i+1;
           }
       }
    }
    printf("%I64d\n",ans);
}