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 N = 2e5 + 1;

int a[N], b[N], c[N];
int lb[N], rb[N];

int main () {
	int n;
	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);

	for (int i = 1; i <= n; i ++)
		rb[a[i]] = i;
	for (int i = n; i >= 1; i --)
		lb[b[i]] = i;

	int sh = 0;
	for (int i = 1; i <= n; i ++) if (lb[i] && rb[i])
		sh = max(sh, rb[i] - lb[i] + 1);

	for (int i = 1; i <= n; i ++)
		c[(i + sh > n) ? (i + sh - n) : (i + sh)] = b[i];

	int ok = 1;
	for (int i = 1; i <= n; i ++) if (a[i] == c[i])
		ok = 0;

	if (ok) {
		puts("Yes");
		for (int i = 1; i <= n; i ++)
			printf("%d%c", c[i], "\n "[i != n]);
	} else {
		puts("No");
	}
}
