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 main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    for(int &i : a){
        cin >> i;
    }

    for(int &i : b){
        cin >> i;
    }
    reverse(b.begin(), b.end());
    int c = -1;
    for(int i = 0; i < n; i++){
        if(a[i] == b[i]){
            c = b[i];
            break;
        }
    }
    int j = 0;
    for(int i = 0; i < n; i++){
        while(j < n && (b[j] == c || a[j] == c)){
            j++;
        }
        if(a[i] == b[i]){
            if(j == n){
                cout << "No";
                return 0;
            }
            int t = b[i];
            b[i] = b[j];
            b[j] = t;
        }
    }
    cout << "Yes\n";
    for(int i : b){
        cout << i << " ";
    }
    return 0;
}
