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;
int n,a[200010],b[200010],cnt[200010],L,R,l,r;
inline int read()
{
	int x=0,w=0;char ch=0;
	while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
	while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return w?-x:x;
}
int main()
{
	r=L=n=read();
	for(int i=1;i<=n;i++)cnt[a[i]=read()]++;
	for(int i=1;i<=n;i++)cnt[b[n-i+1]=read()]++;
	for(int i=1;i<=n;i++)if(cnt[i]>n){cout<<"No";return 0;}
	cout<<"Yes\n";
	l=R=1;
	for(int i=1;i<=n;i++)
	if(a[i]==b[i]){L=min(L,i);R=max(R,i);}
	for(int i=L;i<=R;i++)
	if(l<L&&b[l]!=b[i]&&b[i]!=a[l])swap(b[i],b[l++]);
	else swap(b[i],b[r--]);
	for(int i=1;i<=n;i++)printf("%d ",b[i]);
}