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(){
	int N;
	cin >> N;
	vector<int> A(N),B(N);
	for(auto &i:A)cin >> i;
	for(auto &i:B)cin >> i;
	for(auto &i:A)i--;
	for(auto &i:B)i--;
	
	vector<int> Acount(N,0),Bcount(N,0);
	for(auto &i:A)Acount[i]++;
	for(auto &i:B)Bcount[i]++;
	
	int maxsum=0,maxi;
	for(int i=0;i<N;i++){
		if(maxsum<Acount[i]+Bcount[i]){
			maxsum=Acount[i]+Bcount[i];
			maxi=i;
		}
	}
	if(maxsum>N){
		cout << "No" << endl;
		return 0;
	}
	
	cout << "Yes" << endl;
	int diff=-N;
	for(int i=0;i<N;i++){
		if(Acount[i]+Bcount[i]==0)continue;
		
		int Apos=(int)(lower_bound(A.begin(),A.end(),i)-A.begin());
		int Bpos=(int)(lower_bound(B.begin(),B.end(),i)-B.begin());
		
		diff=max(diff,Apos-Bpos+Acount[i]);
	}
	
	for(int i=0;i<N;i++)printf("%d%c",B[(i-diff+N)%N]+1,((i==N-1)?('\n'):(' ')));
	return 0;
}