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"
using namespace std;

int main() {
	int N;
	cin >> N;
	vector<int>A(N);
	vector<int>B(N);
	vector<int>C(N + 1);
	vector<int>D(N + 1);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		C[A[i]]++;
	}
	for (int i = 0; i < N; ++i) {
		cin >> B[i];
		D[B[i]]++;
	}
	for (int i = 1; i <= N; ++i) {
		if (C[i] + D[i] > N) {
			cout << "No" << endl;
			return 0;
		}
	}
	for (int i = 1; i <= N; i++) {
		C[i] += C[i - 1];
		D[i] += D[i - 1];
	}
	int x = -1;
	for (int i = 1; i <= N; ++i) {
		x = max(x, C[i] - D[i - 1]);
	}
	cout << "Yes" << endl;
	for (int i = 0; i < N; ++i) {
		if (0 != i) {
			cout << " ";
		}
		cout << B[(i + N - x)%N];
	}
	cout << endl;
	return 0;
}
