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
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const double eps=1e-6,pi=acos(-1);
const int N=2e5+5,M=1e9+7,SEGM=4*N,OO=0x3f3f3f3f;
int t,n,m,k;
int a[N];
int b[N];

int main(){
    //freopen("myfile.txt","w",stdout);
    scanf("%d",&n);
    for(int i=0;i<n;++i)
        scanf("%d",a+i);
    for(int i=n-1;i>-1;--i)
        scanf("%d",b+i);
    int l=n,r=-1;
    for(int i=0;i<n;++i){
        if(a[i]==b[i]){
            l=i;
            break;
        }
    }
    for(int i=n-1;i>-1;--i){
        if(a[i]==b[i]){
            r=i;
            break;
        }
    }
    for(int i=0;i<n;++i)
        if(l<=r&&b[l]!=a[i]&&b[l]!=b[i])
            swap(b[i],b[l++]);
    if(l<=r){
        printf("No\n");
    }else{
        printf("Yes\n");
        for(int i=0;i<n;++i)
            printf("%d%c",b[i]," \n"[i+1==n]);
    }
    return 0;
}
