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.
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.
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.
3 2 3 2 2 3 2 2 3
2
6 6 3 1 1 3 7 5 1 2 3 4 5 2 3 4 5 1
1
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 two scientists will be very pleased and all the others will be not satisfied.
/*
这道题可以用离散化来做, 但是仔细分析一下其实也不需要
我们需要遍历每一部电影, 计算有多少人会pleased, 有多少人对satisfied, 然后选出最优电影.
我们将每个scientist会的语言放入到person数组, 将每部电影的音屏和字幕放入audio, subtitle数组.
如何计算每部电影的pleased和satisfied呢?
我们只需要计算这个语言在person数组中出现的个数即可.
而将person数组排序后, 利用upper_bound和lower_bound即可快速在log(n)的时间中求出来
*/
#include <iostream>
#include <algorithm>
#include <unordered_set>
#include <unordered_map>
#include <vector>
#include <cstdlib>
#include <cstdio>
#include <cmath>
using namespace std;
const int MAXN = 1e6;
int subtitle[MAXN], audio[MAXN], person[MAXN];
int n, m;
int main(){
scanf("%d", &n);
for(int i = 0; i < n; ++i){
scanf("%d", person + i);
}
sort(person, person + n);
scanf("%d", &m);
for(int i = 0; i < m; ++i){
scanf("%d", audio + i);
}
for(int i = 0; i < m; ++i){
scanf("%d", subtitle + i);
}
int pleased = -1, satisfied = -1, bestMovie = -1;
for(int i = 0; i < m; ++i){// 总体算法的时间复杂度为 n*log(n)
// 常数可能比较大, 但却是妥妥的log(n)级
int curPleased = upper_bound(person, person + n, audio[i]) - lower_bound(person, person + n, audio[i]);
int curSatisfied = upper_bound(person, person + n, subtitle[i]) - lower_bound(person, person + n, subtitle[i]);
if(curPleased > pleased){
bestMovie = i + 1;
pleased = curPleased;
satisfied = curSatisfied;
}
else if(curPleased == pleased){
if(curSatisfied > satisfied){
bestMovie = i + 1;
pleased = curPleased;
satisfied = curSatisfied;
}
}
}
printf("%d\n", bestMovie);
return 0;
}