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
#pragma GCC optimize ("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
#include"bits/stdc++.h"
using namespace std;
typedef long long ll;
#define int ll
#define all(x) x.begin(), x.end()
#define trav(i,a) for(auto &i:a) 
inline int in(){int x;scanf("%lld",&x);return x;}
int32_t main()
{
	int n=in();
	vector<int> a(n),b(n);
	for(int &i:a)i=in();for(int &i:b)i=in();
	reverse(all(b));
	int c=-1;
	for(int i=0;i<n;i++)
	if(a[i]==b[i]){c=a[i];break;}
	int l=0,r=-1;
	for(int i=0;i<n;i++)
	{
		if(b[i]==c and a[i]==c){l=i;break;}
	}
	for(int i=n-1;i>=0;i--)
	if(b[i]==c and a[i]==c){r=i;break;}
	
	for(int i=0;i<n;i++)
	{
		if(a[i]!=c and b[i]!=c and l<=r)
		{swap(b[i],b[l]);l++;}
	}
	if(l<=r){puts("No");return 0;}
	puts("Yes");
	for(int i:b)cout<<i<<" ";
	
	
}
