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>
#define rep(i,n) for(int i = 0; i < (int)(n); ++i)

using namespace std;
using ll = long long;

int main() {
	int n;
	cin >> n;
	vector<int> cnt(n + 1), a(n), b(n);
	rep(i,n) {
		cin >> a[i];
		cnt[a[i]]++;
	}
	rep(i,n) {
		cin >> b[i];
		cnt[b[i]]++;
	}
	rep(i, n) {
		if (cnt[i + 1] > n) {
			cout << "No" << endl;
			return 0;
		}
	}
	int j = 0;
	rep(i, n) {
		if (a[i] == b[i]) {
			for (; j < n; j++) {
				if (a[j] != a[i] && b[j] != a[i]) {
					swap(b[i], b[j]);
					break;
				}
			}
		}
		if (i != n - 1) {
			if (a[i] != a[i + 1]) j = 0;
		}
	}
	cout << "Yes" << endl;
	rep (i, n) cout << b[i] << ' ';

	return 0;
}
