Problem: 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

Solution (in CPP):

Solution:
#import"bits/stdc++.h"
using namespace std;
#define o bool
#define l int
#define v vector<l>
l n;o h(v a,v b){o r=1;for(l i=0;i<n;i++)r&=(a[i]!=b[i]);return r;}void f(o g,v a){cout<<(g?"Yes\n":"No\n");if(g)for(l e:a)cout<<e<<'\n';exit(0);}l main(){cin>>n;v a(n),b(n),c;for(l&e:a)cin>>e;for(l&e:b)cin>>e;c=b;reverse(begin(b),end(b));rotate(begin(c),begin(c)+(n/2),end(c));if(h(a,b))f(1,b);if(h(a,c))f(1,c);f(0,a);}