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 n, a[200000], b[200000];

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	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);
	int l = -1, r = -1;
	for(int i = 0; i < n; i++)
		if(a[i] == b[i]){
			int nl = i, nr = i - 1;
			for(int j = i; a[j] == b[j] && j < n; j++) nr++;
			l = nl, r = nr;
			if(r - l + 1 == n){
				cout << "No\n";
				return 0;
			}
			break;
		}
	if(l != -1){
		int j = l;
		for(int i = 0; i < l && j <= r; i++)
			if(a[i] != b[j] && a[j] != b[i])
				swap(b[i], b[j++]);
		for(int i = r + 1; i < n && j <= r; i++)
			if(a[i] != b[j] && a[j] != b[i])
				swap(b[i], b[j++]);
		if(j <= r){
			cout << "No\n";
			return 0;
		}
	}
	cout << "Yes\n";
	for(int i = 0; i < n; i++)
		cout << b[i] << " \n"[i == n - 1];
	return 0;
}