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>
#define rep(i,n) for(int i=0; i<n; i++)
#define PI 3.14159265359
#define INF 1000100100000000000
#define MOD 1000000007
#define all(x) (x).begin(),(x).end()
typedef long long ll;
#define P pair<int, int>
#define PP pair<P,P>
#define T tuple<int,int,int> 
using namespace std;

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

    rep(i,n+1){
      if(ac[i]+bc[i]>n){
        cout << "No" << endl;
        return 0;
      }
    }

    rep(i,n){
        ac[i+1]+=ac[i];
        bc[i+1]+=bc[i];
    }

    int x=0;
    rep(i,n){
        x=max(x,ac[i+1]-bc[i]);
    }

    cout << "Yes" << endl;
    rep(i,n){
        printf("%d ",b[(i+n-x)%n]);
    }

return 0;
}
