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 int long long
#define fi first
#define se second
#define vll vector<int>
#define pii pair<int,int>
#define pb push_back
#define sz(v) (int)(v).size()
#define inf 1e18
#define md 1000000007
#define all(v) (v).begin(),(v).end()
#define rep(i,a,b) for(int i=a;i<b;++i)
using namespace std;

#define M 100010


int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
   	
   	int n;cin>>n;
   	int a[n],b[n];
   	rep(i,0,n)cin>>a[i];
   	rep(i,0,n)cin>>b[i];
   	int i=0,j=0;
   	while(i<n){
   		if(a[i]==b[i]){
   			int k=n;
   			while((a[i]==b[j] or a[j]==b[i]) and k--)j=(j+1)%n;
   			swap(b[i],b[j]);
   			if(a[i]==b[i]){
   				cout<<"No";
   				return 0;
   			}
   		}++i;
   	}
   	cout<<"Yes\n";
   	rep(i,0,n)cout<<b[i]<<" ";
   	
   	
    return 0;
}