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 = 2e5 + 5;
int n,a[N],b[N],pos,pos2;
vector<int> v;
inline bool cmp(const int &i,const int &j) {
	return i > j;
}
signed main() {
	scanf("%lld",&n);
	for(int i = 1;i <= n;++i)
		scanf("%d",a+i);
	for(int i = 1;i <= n;++i)
		scanf("%d",b+i);
	sort(b+1,b+n+1,cmp);
	for(int i = 1;i <= n;++i)
		if(a[i] == b[i]) {
			pos = i;
			break;
		}
	if(!pos) {
		puts("Yes");
		for(int i = 1;i <= n;++i)
			printf("%d ",b[i]);
		return puts(""),0;
	}
	for(int i = 1;i <= n;++i)
		if(a[i] != a[pos] && b[i] != a[pos])
			v.push_back(i);
		else if(a[i] == b[i]) pos2 = i;
	if(v.size() >= pos2-pos+1) {
		puts("Yes");
		int cur = 0;
		for(int i = pos;i <= pos2;++i)
			swap(b[i],b[v[cur++]]);
		for(int i = 1;i <= n;++i)
			printf("%d ",b[i]);
		return puts(""),0;
	}
	else puts("No");
	return 0;
}
