codeforces814b

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

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.

题解：

因为必定有解

只会有一个或两个位置不同

如果一个位置不同 留一个合法的不变 改另一个不合法的  使序列变成两位置不同

如果两位置不同 那么c一个不同的位置取a，另一个取b就得到了c

代码：

#include<cstdio>
using namespace std;
const int maxn=1010;
int a[maxn],b[maxn],c[maxn],vis[maxn];
int main(){
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);
    int p=0,q=0;
    for(int i=1;i<=n;i++)
        if(a[i]!=b[i]){
            if(!p)
                p=i;
            else q=i;
        }
	for(int i=1;i<=n;i++)
		c[i]=a[i];
    if(!q){
        for(int i=1;i<=n;i++)
            vis[a[i]]=1;
        for(int i=1;i<=n;i++)
            if(!vis[i])
                c[p]=i;
    }
    else{
        for(int i=1;i<=n;i++)
            if(i!=p&&i!=q)
                vis[a[i]]=1;
        for(int i=1;i<=n;i++)
            if(i!=p&&i!=q)
                vis[b[i]]=1;
        if(!vis[a[p]]&&vis[b[p]]) 
			c[p]=a[p],vis[c[p]]=1;
        if(!vis[a[q]]&&vis[b[q]]) 
			c[q]=a[q],vis[c[q]]=1;
        if(vis[a[q]]&&!vis[b[q]]) 
			c[q]=b[q],vis[c[q]]=1;
        if(vis[a[p]]&&!vis[b[p]]) 
			c[p]=b[p],vis[c[p]]=1;
        if(a[p]==a[q]&&b[p]==b[q])
            c[q]=b[q];
    }
    for(int i=1;i<=n;i++)
        printf("%d ",c[i]);
return 0;
}