Codeforces Round #418 (Div. 2) B. An express train to reveries

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.

解体思路：

由题可知，p中不能有重复的，而且数据必合理，那么只有两种情况：

1. 只在一个下标上不同，这时候只要将a[i]此时的值替换为之前从未在a数组中出现的数字便可；

2.在两个下标上不同，只需交换一次ab即可满足，这时必有两种情况:

(1).a第一个下标代表的数是重复的，此时b必不重复,那么就和b交换,但这时有个特殊情况那就是，和第一个下标重复的正好是第二个下标，进行判断：

如果第一个下标下的代表的数字在a中没出现过那么就交换第一个下标的ab,否则交换第二个下标中的ab；

(2).若a第一个下标代表的数不是重复的,此时只需交换第二个下标即可；

实现代码:

#include<bits/stdc++.h>
using namespace std;

int main()
{
    int t,i,a[1009],b[1009],c[1009],flag[1009];
    cin>>t;
    for(i=1;i<=t;i++)
        flag[i] = 0;
    for(i=0;i<t;i++){
        cin>>a[i];
        flag[a[i]]++;
    }
    for(i=0;i<t;i++)
        cin>>b[i];
    int cnt = 0;
    for(i=0;i<t;i++){
        if(a[i]!=b[i]){
            c[cnt] = i;
            cnt++;
        }
    }
    if(cnt==1){
        for(i=1;i<=t;i++){
            if(flag[i]==0){
                a[c[0]] = i;
            }
        }
    }
    else{if(flag[a[c[0]]]>1&&a[c[0]]!=a[c[1]])
                a[c[0]] = b[c[0]];
            else if(flag[a[c[0]]]>1&&a[c[0]]==a[c[1]]){
                if(flag[b[c[0]]]==0)
                    a[c[0]] = b[c[0]];
                else
                    a[c[1]] = b[c[1]];
            }
            else
                a[c[1]] = b[c[1]];
        }
    for(i=0;i<t-1;i++)
        cout<<a[i]<<" ";
    cout<<a[t-1]<<endl;
    return 0;
}