Problem: Given are two sequences A and B, both of length N. A and B are each sorted in the ascending order. Check if it is possible to reorder the terms of B so that for each i (1 \leq i \leq N) A_i \neq B_i holds, and if it is possible, output any of the reorderings that achieve it.

Constraints

* 1\leq N \leq 2 \times 10^5
* 1\leq A_i,B_i \leq N
* A and B are each sorted in the ascending order.
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 A_2 \cdots A_N
B_1 B_2 \cdots B_N


Output

If there exist no reorderings that satisfy the condition, print `No`.

If there exists a reordering that satisfies the condition, print `Yes` on the first line. After that, print a reordering of B on the second line, separating terms with a whitespace.

If there are multiple reorderings that satisfy the condition, you can print any of them.

Examples

Input

6
1 1 1 2 2 3
1 1 1 2 2 3


Output

Yes
2 2 3 1 1 1


Input

3
1 1 2
1 1 3


Output

No


Input

4
1 1 2 3
1 2 3 3


Output

Yes
3 3 1 2

Solution (in CPP):

Solution:
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn=2e5+10;
int a[maxn],b[maxn];//,c[maxn];
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int n;cin>>n;
    for(int i=1; i<=n; i++)cin>>a[i];
    for(int i=1; i<=n; i++)cin>>b[i];
    int r=n,l;
    reverse(b+1,b+1+n);//for(int i=1; i<=n; i++)c[i]=b[r--];
    l=1;r=n;
    for(int i=1; i<=n; i++){
        if(a[i]!=b[i])continue;
        if(l<i&&b[l]!=a[i]&&a[l]!=b[i]){
            swap(b[l],b[i]);
            l++;
        }else if(i<r&&b[r]!=a[i]&&a[r]!=b[i]){
            swap(b[r],b[i]);
            r--;
        }else return cout<<"No"<<endl,0;
    }
    cout<<"Yes"<<endl;
    for(int i=1; i<=n; i++)cout<<b[i]<<' ';
    return 0;
}