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 REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1e9
#define LINF 1e18
typedef long long ll;

int main() {
    int N; cin >> N;
    vector<int> A(N,0), B(N,0);

    REP(i,N) cin >> A[i];
    REP(i,N) cin >> B[i];
    sort(ALL(B), greater<int>());

    int itr = 0;
    REP(i,N){
        if(A[i] == B[i]){
            while(A[itr]==B[i] || B[itr]==B[i]){
                if(itr == N-1){
                    cout << "No" << endl;
                    return 0;
                }
                itr++;
            }
            swap(B[i],B[itr]);
        }
    }
    cout << "Yes" << endl;
    REP(i,N) cout << B[i] << " ";
    cout << endl;
    return 0;
}