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;

typedef long long LL;

const int N=200000;

int n,a[N+9],b[N+9],p[2][N+9],ans[N+9];

int main(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
	scanf("%d",&a[i]);
  for (int i=n;i>=1;--i)
	if (!p[0][a[i]]) p[0][a[i]]=i;
  for (int i=1;i<=n;++i){
	scanf("%d",&b[i]);
	if (!p[1][b[i]]) p[1][b[i]]=i;
  }
  int mx=0;
  for (int i=1;i<=n;++i)
	if (p[0][i]&&p[1][i]) mx=max(mx,p[0][i]-p[1][i]+1);
  for (int i=1;i<=n;++i) ans[(i+mx-1)%n+1]=b[i];
  for (int i=1;i<=n;++i)
	if (a[i]==ans[i]) {puts("No");return 0;}
  puts("Yes");
  for (int i=1;i<=n;++i)
	printf("%d ",ans[i]);
  puts("");
  return 0;
}