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;
#define ep emplace_back
int main(){
	int n;
	scanf("%d", &n);
	int a[n], b[n], x = -1, c = 0, p = 0;
	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);
	for (int i = 0; i < n; i++) if (a[i] == b[i]) x = a[i], c++;
	if (x == -1){
		printf("Yes\n");
		for (int i = 0; i < n; i++) printf("%d ", b[i]);
		printf("\n");
		return 0;
	}
	vector<int> v;
	for (int i = 0; i < n; i++) if (a[i] != x && b[i] != x) v.ep(i);
	if (c > v.size()){
		printf("No\n");
		return 0;
	}
	for (int i = 0; i < n; i++){
		if (a[i] == b[i]) swap(b[i], b[v[p++]]);
	}
	printf("Yes\n");
	for (int i = 0; i < n; i++) printf("%d ", b[i]);
	printf("\n");
}