B. An express train to reveries

B. An express train to reveries

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Sengoku still remembers the mysterious "colourful meteoroids" she discovered with Lala-chan when they were little. In particular, one of the nights impressed her deeply, giving her the illusion that all her fancies would be realized.

On that night, Sengoku constructed a permutation p1, p2, ..., pn of integers from 1 to n inclusive, with each integer representing a colour, wishing for the colours to see in the coming meteor outburst. Two incredible outbursts then arrived, each with n meteorids, colours of which being integer sequences a1, a2, ..., an and b1, b2, ..., bn respectively. Meteoroids' colours were also between 1 and n inclusive, and the two sequences were not identical, that is, at least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.

Well, she almost had it all — each of the sequences a and b matched exactly n - 1 elements in Sengoku's permutation. In other words, there is exactly one i (1 ≤ i ≤ n) such that ai ≠ pi, and exactly one j (1 ≤ j ≤ n) such that bj ≠ pj.

For now, Sengoku is able to recover the actual colour sequences a and b through astronomical records, but her wishes have been long forgotten. You are to reconstruct any possible permutation Sengoku could have had on that night.

Input

The first line of input contains a positive integer n (2 ≤ n ≤ 1 000) — the length of Sengoku's permutation, being the length of both meteor outbursts at the same time.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ n) — the sequence of colours in the first meteor outburst.

The third line contains n space-separated integers b1, b2, ..., bn (1 ≤ bi ≤ n) — the sequence of colours in the second meteor outburst. At least one i (1 ≤ i ≤ n) exists, such that ai ≠ bi holds.

Output

Output n space-separated integers p1, p2, ..., pn, denoting a possible permutation Sengoku could have had. If there are more than one possible answer, output any one of them.

Input guarantees that such permutation exists.

Examples

input

5
1 2 3 4 3
1 2 5 4 5

output

1 2 5 4 3

input

5
4 4 2 3 1
5 4 5 3 1

output

5 4 2 3 1

input

4
1 1 3 4
1 4 3 4

output

1 2 3 4

Note

In the first sample, both 1, 2, 5, 4, 3 and 1, 2, 3, 4, 5 are acceptable outputs.

In the second sample, 5, 4, 2, 3, 1 is the only permutation to satisfy the constraints.

/*----------------------------------------------
File: F:ACM源代码code forces418B.cpp
Date: 2017/6/8 17:17:55
Author: LyuCheng
----------------------------------------------*/
/*
题意：给你序列a,b，让你构造一个1到n的全排列序列p，使得恰好有一个i使得ai!=pi,恰好有一个j使得bj!==pj
    给出的样例保证有解
思路：考虑下来，a,b不相等的位置最多有两处，并且a,b序列中出现过的数一定是1-n中的n-1个
    当有一处不同的时候：
        直接将a中两个重复数字的其中一个数字换成没有出现过的那个数字，然后输出

    当有两处不同的时候：
        这种情况有很多种，但是简单粗暴的办法就是，先求出，在a,b序列中分别没出现过的数，有两个，然后
        a[i]==b[i]的地方，p[i]=a[i],不相等的地方，先假定p[a,b不相等位置1]=tmp1,p[a,b不相等位置2]=tmp2,
        然后加一个判断是否满足，输出的条件：恰好有一个i使得ai!=pi,恰好有一个j使得bj!==pj,如果不满足，就挑换位置
*/
#include <bits/stdc++.h>
#define MAXN 1005
#define LL long long
using namespace std;

int n;
int a[MAXN];
int visa[MAXN];
int b[MAXN];
int visb[MAXN];
int c[MAXN];
int main(int argc, char *argv[])
{
    // freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        visa[a[i]]=true;
    }
    for(int i=0;i<n;i++){
        scanf("%d",&b[i]);
        visb[b[i]]=true;
    }
    vector<int>v;
    for(int i=0;i<n;i++){
        if(a[i]==b[i]){
            c[i]=a[i];
        }else{
            c[i]=0;
            v.push_back(i);
        }
    }
    if(v.size()==1){//只有一个的时候
        for(int i=1;i<=n;i++){
            if(visa[i]==false&&visb[i]==false){
                c[v[0]]=i;
                break;
            }
        }
    }else{//有两个位置不同的时候
        int tmp1,tmp2;
        for(int i=1;i<=n;i++){
            if(visa[i]==false) tmp1=i;
            if(visb[i]==false) tmp2=i;
        }
        c[v[0]]=tmp1;
        c[v[1]]=tmp2;
        int cur=0;
        for(int i=0;i<n;i++){
            if(a[i]!=c[i]){
                cur++;
            }
        }
        for(int i=0;i<n;i++){
            if(b[i]!=c[i]){
                cur++;
            }
        }
        if(cur>2){
            swap(c[v[0]],c[v[1]]);
        }
    }
    for(int i=0;i<n;i++){
        printf(i==0?"%d":" %d",c[i]);
    }
    printf("
");
    return 0;
}