An express train to reveries

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：

题目描述有一点恶心，先讲一讲题意。

说白了就是给你两个数列a和b，要你找一个数列c，使得c与a和b都最多只有一个不同的数，这就是为什么第二组样例只能有一组解的原因。

思路就是一个一个找a和b相同的数直接放到c中，然后分别试一试两种情况就可以了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<queue>
#include<ctime>
#include<stack>
#include<vector>
using namespace std;
int n,a[10001],b[10001],c[10001],vis[10001];
int cnt1,cnt2,cnt3,cnt4;
int main()
{
int i,j;
scanf("%d",&n);
for(i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}
for(i=1; i<=n; i++)
{
scanf("%d",&b[i]);
}
memset(c,-1,sizeof(c));
for(i=1; i<=n; i++)
{
if(a[i]==b[i])
{
if(!vis[a[i]])
{
c[i]=a[i];
vis[a[i]]=1;
}
}
}
for(i=1; i<=n; i++)
{
if(c[i]==-1)
{
if(!cnt1)cnt1=i;
else
{
cnt2=i;
break;
}
}
}
for(i=1; i<=n; i++)
{
if(!vis[i])
{
if(!cnt3)cnt3=i;
else
{
cnt4=i;
break;
}
}
}
if(!cnt2)c[cnt1]=cnt3;
else
{
int ans1=0,ans2=0;
if(a[cnt1]!=cnt3)ans1++;
if(b[cnt1]!=cnt3)ans1++;
if(a[cnt2]!=cnt4)ans2++;
if(b[cnt2]!=cnt4)ans2++;
if(ans1==1&&ans2==1)
{
c[cnt1]=cnt3;
c[cnt2]=cnt4;
}
else
{
c[cnt2]=cnt3;
c[cnt1]=cnt4;
}
}

for(i=1; i<=n; i++)
cout<<c[i]<<' ';
return 0;
}