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 <iostream>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#define rep(i, n) for(int i = 0; i < (n); i++)
using namespace std;
using P = pair<int, int>;
long mod = (long) 1e9 + 7;

int main(){
    
    int n;
    cin >> n;
    vector<int> a(n,0), b(n,0), c(n+1,0), d(n+1,0);
    rep(i,n){
        cin >> a[i];
        c[a[i]]++;
    }
    rep(i,n){
        cin >> b[i];
        d[b[i]]++;
    }

    int x = 0;
    rep(i,n+1) x = max(x, c[i] + d[i]);
    if(x > n){
        cout << "No" << "\n";
        return 0;
    }
    
    rep(i,n){
        c[i+1] += c[i];
        d[i+1] += d[i];
    }
    
    x = 0;
    rep(i,n) x = max(x, c[i+1] - d[i]);
    
    cout << "Yes" << "\n";
    
    rep(i,n) cout << b[(i+n-x)%n] << ' ';
    cout << "\n";
    
    return 0;
}

