An express train to reveries

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

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.

题目意思：给两个数组，这两个数组至少一个元素不同，并且只有一个位置的元素与原来的数组不同。输出原数组。

总结;wrong answer了4次，原因是没有考虑到插入后可能会出现2个位置不同。

C++CODE:

   #include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
int a[1001],b[1001];
int c[1001];
bool j[1001];
vector<int> index;
int main()
{
int n;
while(cin>>n)
{
memset(j,0,sizeof(j));
for(int i=0;i<n;i++)
cin>>a[i];
for(int i=0;i<n;i++)
cin>>b[i];
for(int i=0;i<n;i++)
{
if(a[i]==b[i])
{
j[a[i]]=1;
}
}
int i;
for(i=0;i<n;i++){
if(a[i]==b[i])
c[i]=a[i];
else if(j[a[i]]&&j[b[i]])
{
for(int k=1;k<=n;k++)
{
if(j[k]==0)
{
c[i]=k;
index.push_back(i);
break;
}
}
}
else{
if(j[a[i]]==1)
{
c[i]=b[i];
j[b[i]]=1;
index.push_back(i);
}
else
{
c[i]=a[i];
j[a[i]]=1;
index.push_back(i);
}

}
}
int counta=0,countb=0;
for(int i=0;i<index.size();i++)
{
if(a[index[i]]!=c[index[i]])
counta++;
if(b[index[i]]!=c[index[i]])
countb++;

}
if(counta==2||countb==2) swap(c[index[0]],c[index[1]]);

for(int i=0;i<n;i++)
{
printf(i==0?"%d":" %d",c[i]);
}
cout<<endl;
index.clear();
}
return 0;
}