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;
void solve(){
	int n,m;
	cin>>n;
	vector<int>A(n);
	vector<int>B(n);
	vector<int>C(n+1);
	bool p=0;
	for(int i=0;i<n;i++){
	 	cin>>A[i];
	 	C[A[i]]++;
	}
	for(int i=0;i<n;i++) {
		cin>>B[i];
		C[B[i]]++;
		if(C[B[i]]>n) p=1;
	}
	if(p){
		cout<<"No";
		return;
	}
	cout<<"Yes \n";
	int i=n-1;
	int j=n-2;
	while(j>=0){
		while(i>=0 && A[i]!=B[i]) i--;
		if(i<=j) j=i-1;
		while(j>=0 && (B[i]==B[j] or A[j]==A[i])) j--;
		if(j<0) break;
		swap(B[i],B[j]);
		j--;
	}
	// for(i=0;i<n;i++) cout<<B[i]<<" ";
	i=0;
	j=1;
	while(j<n){
		while(i<n && A[i]!=B[i]) i++;
		if(i>=j) j=i+1;
		while(j<n  && (B[i]==B[j] or A[j]==A[i])) j++;
		if(j==n) break;
		swap(B[i],B[j]);
		j++;
	}
	//cout<<"\n";
	for(i=0;i<n;i++) cout<<B[i]<<" ";
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	solve();

}