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,x,y) for(int i=x,i##end=y;i<=i##end;++i)
#define _rep(i,x,y) for(int i=x,i##end=y;i>=i##end;--i)
#define ll long long
const int mod=1e9+7;
const int inf=1e9;
#define N 200005
int n,a[N],b[N],c[N];
bool vis[N];
int fyy[N],val=0;
std::vector<int> same,diff;
int main(){
	std::ios::sync_with_stdio(0);
	std::cin>>n;
	rep(i,1,n) std::cin>>a[i],++fyy[a[i]];
	rep(i,1,n) std::cin>>b[i],++fyy[b[i]];
	std::sort(b+1,b+n+1,std::greater<int>());
	rep(i,1,n) if(fyy[i]>n) return puts("No"),0;
	rep(i,1,n){
		if(a[i]==b[i]) same.push_back(i),val=a[i];
	}
	rep(i,1,n) if(a[i]!=val&&b[i]!=val) diff.push_back(i);
	rep(i,0,same.size()-1) std::swap(b[same[i]],b[diff[i]]);
	std::cout<<"Yes\n";
	rep(i,1,n) std::cout<<b[i]<<' '; std::cout<<'\n';
	return 0;
}