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<cstdio>
#include<algorithm>
using namespace std;
int n,a[200005],c[200005],b[200005],left,right,same;
int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;++i) scanf("%d",&a[i]);
	for(int i=1;i<=n;++i) scanf("%d",&c[i]);
	for(int i=1;i<=n;++i) b[i]=c[n-i+1];
	for(int i=1;i<=n;++i) if(a[i]==b[i]) {
		if(!left) left=i;right=i;same=a[i];
	}
	if(left) {
		for(int i=n;i>=right;--i) {
			if(a[i]==same||b[i]==same) break;
			swap(b[i],b[left]);++left;if(left>right) break;
		}
		if(left<=right)
		for(int i=1;i<=left;++i) {
			if(a[i]==same||b[i]==same) break;
			swap(b[i],b[right]);--right;if(left>right) break;
		}
	}
	if(left>right||!left) {printf("Yes\n");for(int i=1;i<=n;++i) printf("%d ",b[i]);}
	else printf("No");
	return 0;
}