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(){
    int N;
    cin >> N;
    vector<int> A(N), B(N);
    for(int i=0; i<N; i++) cin >> A[i];
    for(int i=0; i<N; i++) cin >> B[i];
    
    reverse(B.begin(), B.end());
    int dup = -1;
    vector<int> dup_idx, avail_idx;
    for(int i=0; i<N; i++) if(A[i] == B[i]){
        dup = A[i];
        dup_idx.push_back(i);
    }
    for(int i=0; i<N; i++) if(A[i] != dup && B[i] != dup) avail_idx.push_back(i);
    
    if(dup_idx.size() > avail_idx.size()){
        cout << "No" << endl;
        return 0;
    }

    cout << "Yes" << endl;
    for(int i=0; i<int(dup_idx.size()); i++) swap(B[dup_idx[i]], B[avail_idx[i]]);
    for(int i=0; i<N; i++) cout << B[i] << " \n"[i==N-1];
    return 0;
}
