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 <bits/stdc++.h>
using namespace std;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	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];
	vector<int> p(n);
	for (int i = 0; i < n; i++) {
		p[i] = upper_bound(b.begin(), b.end(), a[i]) - b.begin();
		if (i - 1 >= 0) p[i] = max(p[i], p[i - 1] + 1);
	}
	for (int i = n - 2; i >= 0; i--) {
		p[i] = max(p[i], p[i + 1] - 1);
	}
	for (int i = 0; i < n; i++) {
		if (a[i] == b[p[i] % n]) {
			cout << "No" << "\n";
			return 0;
		}
	}
	cout << "Yes" << "\n";
	for (int i = 0; i < n; i++) {
		cout << b[p[i] % n] << " \n"[i == n - 1];
	}
	return 0;
}