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;

const int SIZE = 2e5+10;
int n;
int a[SIZE], b[SIZE], l[SIZE], r[SIZE];

inline int read()
{	
	int x=0;char ch=getchar();
	while(!isdigit(ch)) ch=getchar();
	while(isdigit(ch)){x=x*10+(ch-'0'); ch=getchar();}
	return x;
}

int main()
{
	int l = SIZE, r = 0, c = 0;
	n=read();
	for (int i = 1; i <= n; ++i) a[i]=read();
	for (int j = 1; j <= n; ++j) b[j]=read();
	for (int i = 1; i <= (n>>1);++i) swap(b[i], b[n-i+1]);
	for (int i = 1; i <= n; ++i)
		if (a[i] == b[i]) { r = i; if (l > n) l = i;}
	for (int i = 1; i <= n; ++i)
		if (a[i] != a[r] && b[i] != a[r]) ++c;
	if (c < r - l + 1) printf("No\n");
	else
	{
		printf("Yes\n");
		for (int i = 1; i <= n; ++i)
		{
			if (l > r) break;
			if (a[i] != a[l] && b[i] != a[l]) swap(b[i], b[l++]);
		}
		for (int i = 1; i <= n; ++i) printf("%d ", b[i]);
	}
	return 0;
}