CodeForces

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

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

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.

题意：一个数字n和两个长度为n数组，这两个数组都只有一个数是错误的。正确的字符数组只包含1~n且每个数出现一次。输出正确的数组。

思路：因为每个数组只有一个数错误，所以我们选择一个数组进行修改，一个数出现两次，或一次没有出现就要修改。但是如果有一位上两数组的数相同那么这一位一定

   正确。

#include<stdio.h>#include<map>#define maxn 1010using namespace std;int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int a[maxn],b[maxn],i,j,t=0;        map<int,int> m1;        map<int,int> m2;        map<int,int> m;        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            m1[a[i]]++;//记录第一个数组数字出现的次数        }        for(i=0;i<n;i++)        {            scanf("%d",&b[i]);            m2[b[i]]++;//记录第二个数组数字出现的次数        }        for(i=1;i<=n;i++)        {            if(m1[i]==0&&m2[i]==0)//看是否有没有出现过的数字            {                t=i;                break;            }        }        for(i=0;i<n;i++)        {            if(m1[a[i]]==2&&a[i]!=b[i])//如果一个数在第一个数组出现两次且这一位上两个数字不同            {                if(m1[b[i]]==0)//在第一个数组中没有这个数                {                    a[i]=b[i];                    m1[a[i]]++;                }                else if(t!=0)//两个数组都没有这个数                    a[i]=t;            }        }        for(i=0;i<n;i++)        {            if(i==n-1)                printf("%d\n",a[i]);            else                printf("%d ",a[i]);        }    }}