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

#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define deb(x)	cout << #x << " " << x << endl;

typedef long long int ll;

int main() {

	IOS

	int n;
	cin >> n;

	int a[n], b[n];

	for (int i = 0; i < n; i++)
		cin >> a[i];

	for (int i = n-1; i >= 0; i--)
		cin >> b[i];

	bool swapped = true;
	int prev = -1, j;

	for (int i = 0; i < n; i++) {
		if (a[i] == b[i]) {
			swapped = false;

			if (a[i] != prev)
				j = 0;

			for (; j < n; j++) {
				if (b[i] == b[j])
					continue;

				if (a[j] == b[i])
					continue;

				if (a[i] == b[j])
					continue;

				swap(b[i], b[j]);
				swapped = true;
				break;
			}

			if (!swapped)
				break;

			prev = a[i];
		}
	}

	if (swapped) {
		cout << "Yes" << endl;
		for (int i = 0; i < n; i++)
			cout << b[i] << " ";

	} else
		cout << "No";

	return 0;
}