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 Maxn 200010
 
int n, a[Maxn], b[Maxn], c[Maxn], P[Maxn];
set<int >S;
int main()
{
	scanf("%d", &n);
	for (int i=1; i<=n; i++)
	{
		scanf("%d", &a[i]);
		P[a[i]]++;
		S.insert(a[i]);
	}
	for (int i=1; i<=n; i++)
	{
		scanf("%d", &b[i]);
		P[b[i]]++;
		S.insert(b[i]);
	}
	bool flag = 0; for (auto p:S) if (P[p]>n) { flag = 1; break; }
	if (flag) puts("No");
	else
	{
		puts("Yes");
		for (int i=1; i<=n; i++) c[i]=b[n-i+1];
		int cur = n, top = 1;
		for (int i=1; i<=n; i++)
			while ((a[i]==c[i]) && (cur >= top))
                if(c[cur] != c[i] && a[cur] != c[i]){ swap(c[i], c[cur]);}
                else if(c[top] != c[i] && a[top] != c[i]) {swap(c[i], c[top]);}
                else cur--, top++;
		for (int i=1; i<=n; i++) printf("%d ", c[i]); puts("");
	}
}