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;
 
#define ll long long
#define ld long double
#define pb push_back
 
#define mo 1000000007
#define inf 1e18
#define rep(i, s, n) for (ll i = s; i < n; i = i + 1)
#define rrep(i,s,n) for(ll i=s;i>=n;i--)

ll mod(ll n) { return (n % (ll)mo + (ll)mo)%(ll)mo;}

int main(){
	ll n;
	cin>>n;
	vector<ll> a(n);
	vector<ll> b(n);
	
	rep(i,0,n) cin>>a[i];
	rep(i,0,n) cin>>b[i];
	ll idx = n-1;
	rep(i,0,n){
		
		
		if(a[i]==b[i]){
			if(a[i]==b[idx]) idx = n-1;
			swap(b[i],b[idx]);
			idx--;
			
		}

	}
	
	idx = 0;
	rep(i,0,n){
		
		
		if(a[i]==b[i]){
			if(a[i]==b[idx]) idx = 0;
			swap(b[i],b[idx]);
			idx++;
			
		}
	
	}
	
	
	ll f = 0;
	rep(i,0,n){
		if(a[i]==b[i]) f=1;
	}
	
	if(f) cout<<"No";
	else{
		cout<<"Yes"<<endl;
		rep(i,0,n) cout<<b[i]<<" ";
		
	}
	
	
}