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 <cstring>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
#define maxn 200010
#define pb push_back

int n,A[maxn],B[maxn];
vector<int> a,b;

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",&B[i]);
	reverse(B+1,B+n+1);
	for(int i=1;i<=n;i++)if(A[i]==B[i])a.pb(i);
	if(a.size()){
		for(int i=1;i<=n;i++)if(A[i]!=A[a[0]]&&B[i]!=B[a[0]])b.pb(i);
		if(b.size()<a.size())return printf("No\n"),0;
		for(int i=0;i<a.size();i++)swap(B[b[i]],B[a[i]]);
	}
	printf("Yes\n");
	for(int i=1;i<=n;i++)printf("%d ",B[i]);
}