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;

int main(){
	int n; cin >> n;
	vector<int> a(n), b(n);
	for(int i = 0; i < n; i++) cin >> a[i];
	for(int i = 0; i < n; i++) cin >> b[i];
	reverse(b.begin(), b.end());
	int l = 0, r = -1, c;
	for(int i = 0; i < n; i++)
		if(a[i]==b[i]){
			l = i; c = a[i]; 
			break;
		}
	for(int i = n-1; 0 <= i; i--)
		if(a[i]==b[i] && a[i]==c){
			r = i; break;
		}
	for(int i = 0; i < n && l <= r; i++)
		if(a[i]!=c && b[i]!=c) swap(b[i], b[l++]);
	if(l<=r) cout << "No\n";
	else {
		cout << "Yes\n";
		for(int i = 0; i < n; i++) cout << b[i] << ' ';
		cout << '\n';
	}
	return 0;
}
