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() {
	ios::sync_with_stdio(0);
	cin.tie(0);

	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];
	}
	reverse(b.begin(), b.end());
	int start = -1, fin = -1;
	for (int i = 0; i < n; ++i) {
		if (b[i] == a[i]) {
			if (start == -1) start = i;
			fin = i;
		}
	}
	bool bad = 0;
	if (start != -1) {
		for (int i = 0, j = start; j <= fin; ++i) {
			if (i == start) i = fin + 1;
			if (i >= n) {
				bad = 1;
				break;
			}
			if (a[i] != b[j] && b[i] != b[j]) {
				swap(b[i], b[j++]);
			}
		}
		for (int i = 0; i < n; ++i)
			bad |= (a[i] == b[i]);
	}
	if (bad) {
		cout << "No" << endl;
	}
	else {
		cout << "Yes" << endl;
		for (int i = 0; i < n; ++i) {
			if (i) cout << " ";
			cout << b[i];
		}
		cout << endl;
	}

	return 0;
}