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 = 200005;

int n;
int a[N], b[N];

int main() {
	scanf("%d", &n);

	for(int i = 0; i < n; i++) scanf("%d", a + i);
	for(int i = 0; i < n; i++) scanf("%d", b + i);
	reverse(b, b + n);

	int l = 0;
	while(l < n and a[l] != b[l]) l++;

	if(l < n) {
		int r = l;
		while(r + 1 < n and a[r + 1] == b[r + 1]) r++;
		int pos = l;
		for(int i = 0; i < l and pos <= r; i++) {
			if(a[i] != b[pos] and b[i] != b[pos]) {
				swap(b[i], b[pos++]);
			}
		}
		for(int i = r + 1; i < n and pos <= r; i++) {
			if(a[i] != b[pos] and b[i] != b[pos]) {
				swap(b[i], b[pos++]);
			}
		}
		if(pos <= r) return printf("No\n"), 0;
	}

	printf("Yes\n");
	for(int i = 0; i < n; i++) {
		printf("%d%c", b[i], " \n"[i + 1 == n]);
	}
}
