Problem: 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

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1000000007
using namespace std;

int main(){
  int n;
  cin >> n;
  vector<int> a(n),b(n),asum(n+1,0),bsum(n+1,0);
  rep(i,n){
    cin >> a[i];
    asum[a[i]]++;
  }
  rep(i,n){
    cin >> b[i];
    bsum[b[i]]++;
  }
  rep(i,n+1){
    if(asum[i] + bsum[i] > n){
      cout << "No\n";
      return 0;
    }
  }
  rep(i,n){
    asum[i+1] += asum[i];
    bsum[i+1] += bsum[i];
  }
  int d = 0;
  rep(i,n+1){
    d = max(asum[i]-bsum[i-1],d);
  }
  cout << "Yes\n";
  rep(i,n){
    cout << b[(n+i-d)%n] << " ";
  }
  return 0;
}