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;

#define endl '\n'
#define int long long

int32_t main()
{
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

	int n; cin >> n;
	vector<int> a(n), b(n), cnt(n + 1, 0);
	for(int i = 0; i < n; i ++) {
		cin >> a[i];
		cnt[a[i]] ++;
	}
	for(int i = 0; i < n; i ++) {
		cin >> b[i];
		cnt[b[i]] ++;
	}
	int m = *max_element(cnt.begin(), cnt.end());
	if(m > n) {
		cout << "No" << endl;
	}
	else {
		reverse(b.begin(), b.end());
		int l = 0, r = n - 1;
		for(int i = 0; i < n; i ++) {
			if(a[i] == b[i]) {
				if(a[i] != b[l] and a[l] != b[i]) {
					swap(b[l], b[i]);
					l ++;
				}
				else {
					swap(b[r], b[i]);
					r --;
				}
			}
		}
		cout << "Yes" << endl;
		for(int i : b) cout << i << " ";
	}

	return 0;
}