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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, ..., p_n 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, ..., a_n and b₁, b₂, ..., b_nrespectively. 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 a_i ≠ b_i 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 a_i ≠ p_i, and exactly one j (1 ≤ j ≤ n) such that b_j ≠ p_j.

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, ..., a_n (1 ≤ a_i ≤ n) — the sequence of colours in the first meteor outburst.

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

Output

Output n space-separated integers p₁, p₂, ..., p_n, 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.

Example

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

题目大概：

给出两组序列，输出和两组序列差别都不超过1个数字的序列，并且该序列没有重复数字。

思路：

暴力搜索，模拟，不断更新一个位置数字，直到找到符合条件的序列。

代码：

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int n;int a[1002],q[1002];int b[1002],map[1002];int main(){while(cin>>n){    for(int i=1;i<=n;i++)    {        cin>>a[i];    }    for(int i=1;i<=n;i++)    {        cin>>b[i];    }    for(int i=1;i<=n;i++)    {        memset(map,0,sizeof(map));        int f=0;        for(int j=1;j<=n;j++)        {            if(i==j)continue;            map[a[j]]++;            q[j]=a[j];            if(map[a[j]]>1)f=1;        }        if(f)continue;        for(int j=1;j<=n;j++)        {            if(!map[j]){q[i]=j;break;}        }        int sum=0;        for(int j=1;j<=n;j++)        {            if(b[j]!=q[j])sum++;        }        if(sum==1)break;    }    for(int i=1;i<=n;i++)    {        cout<<q[i];        if(i!=n)cout<<" ";    }    cout<<endl;}    return 0;}