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 <vector>
#include <algorithm>
#include <map>
using namespace std;
//using namespace atcoder;

int main() {
	int N;
	cin >> N;
	vector<int> A(N), B(N);
	map<int, int> ANum, BNum;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		ANum[A[i]]++;
	}
	for (int i = 0; i < N; i++) {
		cin >> B[i];
		BNum[B[i]]++;
	}
	for (int i = 0; i <= N; i++) {
		if (ANum.find(i) == ANum.end() || BNum.find(i) == BNum.end()) continue;
		if (ANum[i] + BNum[i] > N) {
			cout << "No" << endl;
			return 0;
		}
	}
	reverse(B.begin(), B.end());
	int j = N - 1;
	for (int i = 0; i < N; i++) {
		if (A[i] == B[i]) {
			while (A[i] == B[j] || B[i] == A[j]) {
				j--;
				if (j < 0) j = N - 1;
			}
			swap(B[i], B[j]);
		}
	}
	cout << "Yes" << endl;
	for (int b : B) cout << b << " ";
	cout << endl;
}
