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;
#define dlwlrma ios::sync_with_stdio(0); cin.tie(0);
int n;
int main(){
    dlwlrma
    cin >> n;
    vector<int>a(n),b(n);
    for(int &next : a)
        cin >> next;
    for(int &next : b)
        cin >> next;
    reverse(b.begin(),b.end());
    vector<int>same,dif;
    int h;
    for(int i = 0; i < n; i++){
        if(a[i]==b[i])
            h=a[i];
    }
    for(int i = 0; i < n; i++){
        if(a[i]==b[i])
            same.push_back(i);
        if(a[i]!=h&&b[i]!=h)
            dif.push_back(i);
    }
    if(same.size()>dif.size()){
        cout << "No\n";
        return 0;
    }
    cout << "Yes\n";
    for(int i = 0; i < same.size(); i++)
        swap(b[same[i]],b[dif[i]]);
    for(int i = 0; i < n; i++)
        cout << b[i] << " ";
}