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;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
#define F(i,a,n) for(int i=a;i<n;i++)
#define mod 1000000007 
#define pb push_back
int main(){
	#ifndef ONLINE_JUDGE
		freopen("input.txt","r",stdin);
		freopen("output.txt","w",stdout);
	#endif
	int w=1;
	// cin>>w;
	while(w--){
		ll n;
		cin>>n;
		vector<ll>v1(n),v2(n);
		F(i,0,n){
			cin>>v1[i];
		}
		F(i,0,n){
			cin>>v2[i];
		}
		vi v;
		for(int i=0,j=n-1;i<n;i++,j--){
			if(v1[i]==v2[j]){v.pb(j);}
		}
		int x=0;
		for(auto i:v){
			swap(v2[i],v2[x]);
			x++;
		}
		v.clear();
		for(int i=0,j=n-1;i<n;i++,j--){
			if(v1[i]==v2[j]){v.pb(j);}
		}
		x=n-1;
		for(auto i:v){
			swap(v2[i],v2[x]);
			x--;
		}
		for(int i=0,j=n-1;i<n;i++,j--){
			if(v1[i]==v2[j]){cout<<"No";return 0;}
		}
		cout<<"Yes\n";
		for(int j=n-1;j>=0;j--){
			cout<<v2[j]<<" ";
		}

	}
	return 0;
}