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 < n; i++)
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
const int MAX_N = 200005;
int A[MAX_N],B[MAX_N],C[MAX_N],D[MAX_N];
vector<P> v,v2;

int main()
{
	int N;
	cin >> N;
	rep(i,N)
	{
		cin >> A[i];
		C[A[i]]++;
	}
	rep(i,N)
	{
		cin >> B[i];
		D[B[i]]++;
	}
	rep(i,MAX_N)
	{
		if (C[i] + D[i] > N)
		{
			cout << "No" << endl;
			return 0;
		}
	}

	rep(i,N+1)
	{
		C[i+1] += C[i];
		D[i+1] += D[i];
		//cout << C[i+1] << " " << D[i+1] << endl;
	}

	int ans = 0;
	rep(i,N)
	{
		ans = max(ans,C[i+1]-D[i]);
	}
	cout << "Yes" << endl;
	rep(i,N)
	{
		if (i != 0) cout << " ";
		cout << B[(i-ans+N) % N];
	}
	cout << endl;
	return 0;
}