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;

long long int cnt[200005];
long long int cnt2[200005];

int main(void)
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n,t;
	vector <int> a,b;

	cin >> n;

	for(int i=0;i<n;i++)
	{
		cin >> t;
		a.push_back(t);
		cnt[t]++;
	}

	for(int i=0;i<n;i++)
	{
		cin >> t;
		b.push_back(t);
		cnt2[t]++;
	}

	for(int i=1;i<=n;i++)
	{
		if(cnt[i] + cnt2[i] > n)
		{
			cout << "No" << '\n';
			return 0;
		}
	}

	cout << "Yes" << '\n';

	long long int x = 0;

	for(int i=1;i<=n;i++)
	{
		cnt[i] += cnt[i-1];
		cnt2[i] += cnt2[i-1];
		x = max(x,cnt[i] - cnt2[i-1]);
	}

	for(int i=0;i<n;i++)
	{
		int idx = i + n - x;
		idx%=n;
		cout << b[idx] << ' ';
	}
	cout << '\n';

	return 0;
}
