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;
const int maxn = 2e5 + 5;
int a[maxn], b[maxn];
int main(){
    int n;
    cin >> n;
    for(int i = 1; i <= n; i ++) cin >> a[i];
    for(int i = 1; i <= n; i ++) cin >> b[n-i+1];
    int ok = 0;
    for(int i = 1; i <= n; i ++){
        if(a[i] != b[i]) continue;
        int f = 0;
        for(int j = 1; j <= n; j ++){
            if(a[i] != b[j] && a[j] != b[i]){
                f = 1;
                swap(b[i], b[j]);
                break;
            }
        }
        if(!f){
            ok=1;
            goto outer;
        }
    }
	outer:;
    if(ok==1){
        cout << "No" << endl;
    }else{
		cout << "Yes" << endl;
		for(int i = 1; i <= n; i ++){
			if(i != 1)cout << " ";
			cout << b[i];
		}
    	cout << endl;
	}
    return 0;
}
