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>

int main()
{
	int N;
	scanf("%d", &N);
	std::vector<int> A(N), B(N);
	std::vector<std::pair<int, int>> num(N + 1);
	for (int i{}; i <= N; i++)
		num[i].second = i;
	num.reserve(2 * N);
	for (auto& e: A)
	{
		scanf("%d", &e);
		num[e].first++;
	}
	for (auto& e: B)
	{
		scanf("%d", &e);
		num[e].first++;
	}
	std::sort(num.rbegin(), num.rend());
	if (num.front().first > N)
	{
		puts("No");
		return 0;
	}
	puts("Yes");
	std::reverse(B.begin(), B.end());
	
	int change{};
	for (int i{}; i < N; i++)
	{
		if (A[i] != B[i]) continue;
		while (A[change] == A[i] || B[change] == A[i])
			change++;
		std::swap(B[i], B[change]);
		change++;
	}
	for (int i{}; i < N; i++)
		printf("%d%c", B[i], i + 1 == N? '\n': ' ');

	return 0;
}