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;
typedef long long ll;
#define rep(i,a,b) for (ll i = (a); i < (b); i++)
#define REP(i,n) rep(i,0,n)

void solve() 
{
    int n;cin>>n;
    int a[n],b[n];
    REP(i,n)cin>>a[i];
    REP(i,n)cin>>b[i];
    int j=0,now=-1;
    REP(i,n){
        if(now!=a[i])j=0;
        if(a[i]==b[i]){
            for(;j<n;j++){
                if(a[i]!=a[j]&&a[i]!=b[j]){
                    swap(b[i],b[j]);
                    break;
                }
            }
            if(j==n){cout<<"No"<<endl;return;}
        }
        now=a[i];
    }
    cout<<"Yes"<<endl;
    REP(i,n)cout<<(i?" ":"")<<b[i];cout<<endl;
}

int main()
{
    cin.tie(0);
	ios::sync_with_stdio(false);
    solve();
    return 0;
}