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<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<n;i++)
#define N 200010
int n,a[N],b[N],c[N],d[N];
const int INF = 1000000000;
int main(){
    cin>>n;
    rep(i,n)cin>>a[i];
    rep(i,n)cin>>b[i];
    rep(i,n+1)c[i]=d[i]=0;
    rep(i,n)c[a[i]]++,d[b[i]]++;
    rep(i,n+1){
	if(c[i]+d[i]>n){
	    cout<<"No\n";
	    return 0;
	}
    }
    for(int i=1;i<=n;i++){
	c[i]+=c[i-1];
	d[i]+=d[i-1];
    }
    int x=INF;
    for(int i=1;i<=n;i++)x=min(x,c[i-1]+n-d[i]);
    cout<<"Yes\n";
    rep(i,n){
	if(i)cout<<" ";
	cout<<b[(i+n-x)%n];
    }
    cout<<"\n";
}