814/ B. An express train to reveries

题目链接：http://codeforces.com/problemset/problem/814/B

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

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

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.

题意：给你a,b两个串，求p串，要求p串分别和a,b串只有一个不相同，并且p串不能含有重复元素（必须1~n）

解析：a,b不相同个数只能为1或者2，分别处理就好，简单判断下，比赛时代码有点乱，但思路还是比较清晰，不想改了，将就下看吧，哈哈~

代码：

#include<bits/stdc++.h>#define N 2009using namespace std;int a[N], b[N], used[N], p[N];bool judge(int n){    int num1, num2; num1 = num2 = 0;    for(int i = 1; i <= n; i++)    {        if(p[i] != a[i]) num1++;        if(p[i] != b[i]) num2++;    }    if(num1 == num2&&num1 == 1) return true;    return false;}int main(){    int n;    memset(used, 0, sizeof(used));    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 num = 0;    for(int i = 1; i <= n; i++)    {        if(a[i] != b[i]) num++;        else used[a[i]] = 1;    }    if(num == 1)    {        for(int i = 1; i <= n; i++)        {            if(a[i] == b[i]) printf("%d%c", a[i], i == n ? '\n' : ' ');            else            {                for(int j = 1; j <= n; j++)                {                    if(used[j]) continue;                    printf("%d%c", j, j == n ? '\n' : ' ');                    break;                }            }        }    }    else    {        stack<int> s;        for(int i = 1; i <= n; i++) if(!used[i]) s.push(i);        int num1 = s.top(); s.pop();        int num2 = s.top(); s.pop();        while(!s.empty()) s.pop();        for(int i = 1; i <= n; i++)        {            if(a[i] != b[i]) s.push(i);            else p[i] = a[i];        }        int ii = s.top(); s.pop();        int jj = s.top(); s.pop();        while(!s.empty()) s.pop();        p[ii] =  num1; p[jj] = num2;        if(judge(n))        {            for(int i = 1; i <= n; i++) printf("%d%c", p[i], i == n ? '\n' : ' ');        }        else        {            p[ii] =  num2; p[jj] = num1;            for(int i = 1; i <= n; i++) printf("%d%c", p[i], i == n ? '\n' : ' ');        }    }    return 0;}