codeforces 814 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 ninclusive, 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

51 2 3 4 31 2 5 4 5

output

1 2 5 4 3

input

54 4 2 3 15 4 5 3 1

output

5 4 2 3 1

input

41 1 3 41 4 3 4

output

1 2 3 4

题意：最初有一个序列有两个人a和b他们记录这个序列，但他们各自记的序列中分别有一个数记错了，现在给你他们两人的序列，让你给出原序列的一种情况（原序列每个数只出现一次，且都在1到n之间）

判断他们记错的数是否是同一位置，如果同一位置，则直接找到没出现过的数，补上去即可（a，b在同一位置 相同  的数 是出现过的）否则要进行判断

#include <iostream>#include<stdio.h>using namespace std;int main(){    int a[1100],b[1100];    int c[1100]={0};    int n;    int d[5][2];//存a和b出现不同的位置的a[i]和b[i];    int t=0;    scanf("%d",&n);    for(int i=0;i<n;i++)        scanf("%d",&a[i]);    for(int i=0;i<n;i++)    {        scanf("%d",&b[i]);        if(a[i]==b[i])            c[a[i]]++;//记录一定没错的数据        else        d[t][0]=a[i],d[t][1]=b[i],t++;//存d    }    if(t==1)//判断记错的位置是否在同一位置    {        for(int i=1;i<=n;i++)        {            if(c[i]==0)            {                for(int j=0;j<n;j++)                {                    if(a[j]!=b[j])                        a[j]=i;                }            }        }    }    else    {        int h=0;        if(c[d[0][0]]==0&&c[d[1][1]]==0)//因为每人只记错一位所以判断是取哪种情况（要么第一处不同取a的则第二处不同取b的否则第一处不同取b的则第二处不同取a）        {            for(int i=1;i<=n;i++)            {                if(c[i]==0)               {                for(int j=0;j<n;j++)                {                    if(a[j]!=b[j])                    {                        if(h==0)                        a[j]=d[0][0],h++;                        else                        a[j]=d[1][1];                    }                }                break;               }            }        }        else        {           // printf("**\n");            int h=0;            for(int i=1;i<=n;i++)            {                if(c[i]==0)                {                    for(int j=0;j<n;j++)                    {                    if(a[j]!=b[j])                    {                        if(h==0)                        a[j]=d[0][1],h++;                        else                       a[j]=d[1][0];                    }                    }                    break;                }            }        }    }    printf("%d",a[0]);    for(int i=1;i<n;i++)        printf(" %d",a[i]);    printf("\n");    return 0;}