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;
int a[200050],b[200050],n;
int main()
{   
        scanf("%d",&n);
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        for(int i=1;i<=n;i++) scanf("%d",&b[i]);
        reverse(b+1,b+1+n);
        vector<int> p,q;
        int val=0,cnt=0;
        for(int i=1;i<=n;i++)
        {
            if(a[i]!=b[i]) continue;
            cnt++;
            p.push_back(i);
            val=b[i];
        }
        for(int i=1;i<=n;i++)
        {
            if(!cnt) break;
            if(a[i]==b[i]) continue;
            if(a[i]!=val&&b[i]!=val)
            {
                cnt--;
                q.push_back(i);
            }
        }
        if(cnt) return cout<<"No\n",0;
        for(int i=0;i<q.size();i++) swap(b[p[i]],b[q[i]]);
        cout<<"Yes\n";
        for(int i=1;i<=n;i++) cout<<b[i]<<' ';
        cout<<'\n';
        return 0;
}