F - An express train to reveries CodeForces - 814B （简单思维）

原创于 2017-08-25 19:00:41 发布 · 263 阅读

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解决两组序列中各有一个错误元素的问题，通过统计分析找到正确的颜色排列。

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, ..., p_n 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 nmeteorids, colours of which being integer sequences a₁, a₂, ..., a_n and b₁, b₂, ..., b_nrespectively. 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 a_i ≠ b_i holds.

Well, she almost had it all — each of the sequences a and b matched exactly n - 1elements in Sengoku's permutation. In other words, there is exactly one i (1 ≤ i ≤ n) such that a_i ≠ p_i, and exactly one j (1 ≤ j ≤ n) such that b_j ≠ p_j.

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, ..., a_n (1 ≤ a_i ≤ n) — the sequence of colours in the first meteor outburst.

The third line contains n space-separated integers b₁, b₂, ..., b_n (1 ≤ b_i ≤ n) — the sequence of colours in the second meteor outburst. At least one i (1 ≤ i ≤ n) exists, such that a_i ≠ b_i holds.

Output

Output n space-separated integers p₁, p₂, ..., p_n, 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.

Example

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.

题意：两串相同长度的数字，其中每串有一个数字是错的，让你找出对的数字串，所有数字都是小于等于n的。

思路：统计第一个串里的数字个数，再统计一下1到n里哪个数字没有出现，如果第一个串中某个数字出现了两次，那么其中有一个肯定是错的，该位上对的数字就是那个没有出现过的数字了。

#include<bits/stdc++.h>
using namespace std;
#define maxn 1005
int a[maxn],b[maxn];
int num[maxn];
int main()
{
    int n;
    int x;
    scanf("%d",&n);
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++){
        scanf("%d",&a[i]);
        num[a[i]]++;
    }
    for(int j=1;j<=n;j++){
        scanf("%d",&b[j]);
        if(num[j]==0) x=j;
    }
    int temp;
    int f;
    for(int i=1;i<=n;i++){
        if(num[a[i]]==2){
         temp=a[i];
         a[i]=x;
         f=0;
         for(int j=1;j<=n;j++)
            if(a[j]!=b[j]) f++;
            if(f==1){
                for(int k=1;k<=n;k++)
                cout<<a[k]<<" ";
                return 0;
            }
             a[i]=temp;
        }
    }
        cout<<endl;
}