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>
#define maxn 200086

using namespace std;

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

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 + 1 + n);
	int l = n, r = 1;
	for(int i = 1;i <= n;i++){
		if(a[i] == b[i]) l = min(l, i), r = max(r, i);
	}
	for(int i = 1;i <= n && l <= r;i++){
		if(a[i] != a[l] && b[i] != a[l]) swap(b[i], b[l++]);
	}
	if(l <= r) return printf("No"), 0;
	puts("Yes");
	for(int i = 1;i <= n;i++) printf("%d ", b[i]);
}