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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>

using namespace std;

const int maxn = 2e5;

int n, a[maxn + 5], b[maxn + 5];
int l[maxn + 5], r[maxn + 5];
int ans[maxn + 5];

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]);
	for (int i = 1; i <= n; ++i)r[a[i]] = i;
	for (int i = n; i > 0; --i)l[b[i]] = i;
	int res = 0;
	for (int i = 1; i <= n; ++i)if (l[i] && r[i])res = max(res, r[i] - l[i] + 1);
	for (int i = 1; i <= n; ++i) {
		int npos = i + res;
		if (npos > n)npos -= n;
		ans[npos] = b[i];
	}
	for (int i = 1; i <= n; ++i) {
		if (a[i] == ans[i]) {
			printf("No\n");
			return 0;
		}
	}
	printf("Yes\n");
	for (int i = 1; i <= n; ++i)printf("%d%c", ans[i], " \n"[i == n]);
	return 0;
}