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)
#define repit(it, li) for(auto it=li.begin(); it!=li.end(); it++)
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main(){
	int n;
	cin>>n;
	vector<int> a(n), b(n);
	rep(i, n) cin>>a[i];
	rep(i, n) cin>>b[i];

	vector<bool> ac(n, true);
	int l=0, r=0, c=0, d=0;
	for(int i=1; i<=n; i++){
		l=r;
		for(; a[r]==i && r<n; r++);
		c=d;
		for(; b[d]==i && d<n; d++);
		if(l<r && c<d){
			rep(i, (r-l)+(d-c)-1) ac[(l-(d-1)+i+n)%n]=false;
		}
	}

	int e= -1;
	rep(i, n) if(ac[i]) e=i;

	if(e== -1){
		cout<<"No"<<endl;
	}
	else{
		cout<<"Yes"<<endl;
		rep(i, n) cout<<b[(-e+i+n)%n]<<" ";
		cout<<endl;
	}
	return 0;
}
