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;
const int maxn=2e5+5;
int a[maxn],b[maxn];
int main(){
    int n;
    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]);
    reverse(b+1,b+1+n);
    int h1=n,q1=1,mx=-1;
    for(int i=1;i<=n;++i){
        if(a[i]==b[i]){
            mx=a[i];
            h1=i;
            while(a[i]==b[i])i++;
            q1=i-1;
        }
    }
    for(int i=1;i<=n;++i){
        if(a[i]!=mx&&b[i]!=mx&&h1<=q1){
            swap(b[i],b[h1]);
            h1++;
        }
    }
    if(h1<=q1){puts("No");return 0;}
    else {
        puts("Yes");
        for(int i=1;i<n;++i)printf("%d ",b[i]);
        cout<<b[n]<<endl;
    }
    return 0;
}