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 <iostream>
#include <vector>
#include <algorithm>

int main() {
	int n;
	std::cin >> n;
	std::vector<int> a(n), b(n);
	for (int &e : a) std::cin >> e;
	for (int &e : b) std::cin >> e;
	
	auto check = [&] () {
		bool ret = true;
		for (int i = 0; i < n; i++) ret &= (a[i] != b[i]);
		return ret;
	};
	
	std::reverse(b.begin(), b.end());
	
	if (check()) {
		std::cout << "Yes" << '\n';
		for (int e : b) std::cout << e << '\n';
	} else {
		std::reverse(b.begin(), b.end());
		std::rotate(b.begin(), b.begin() + (n / 2), b.end());
		if (check()) {
			std::cout << "Yes" << '\n';
			for (int e : b) std::cout << e << '\n';
		} else {
			std::cout << "No" << '\n';
		}
	}
	
	return 0;
}