Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;

const int maxn=2*1e5+6;
int n, a[maxn], b[maxn];
int l=-1, r=-2, c;

int main(){
	cin >> n;
	for (int i=0;i<n;i++) cin >> a[i];
	for (int i=0;i<n;i++) cin >> b[i];
	reverse(b,b+n);
	for (int i=0;i<n;i++){
		if (a[i]==b[i]){
			if (l==-1) l=i;
			r=i;
			c=a[i];
		}
	}
	int idx=0;
	for (int i=l;i<=r;i++){
		while (idx<n && (b[idx]==c || a[idx]==c)) idx++;
		if (idx==n) break;
		swap(b[i],b[idx]);
		idx++;
	}
	for (int i=0;i<n;i++){
		if (a[i]==b[i]){
			cout << "No\n";
			return 0;
		}
	}
	cout << "Yes\n";
	for (int i=0;i<n;i++) cout << b[i] << " ";
	cout << "\n";
}