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
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
#define fi first
#define se second
#define pb push_back
#define wzh(x) cerr<<#x<<'='<<x<<endl;
int n,a[N],b[N],L[N],R[N],f[N];
int main() {
  ios::sync_with_stdio(false);
  cin>>n;
  set<int>al;
  for(int i=1;i<=n;i++)cin>>a[i],L[a[i]]++;
  for(int i=1;i<=n;i++)cin>>b[i],R[b[i]]++;
  for(int i=1;i<=n;i++)if(L[i]+R[i]>n)return cout<<"No\n",0;
  reverse(b+1,b+1+n);
  cout<<"Yes\n";
  int l=-1;
  vector<int>v;
  for(int i=1;i<=n;i++){
    if(a[i]==b[i]){
      l=a[i];
      v.pb(i);
    }
  }
 // cout<<"ok"<<endl;
  int idx=0;
  for(int i=1;i<=n&&idx<v.size();i++){
    if(a[i]!=b[i] && a[i]!=l && b[i]!=l){
      swap(b[i],b[v[idx++]]);
    }
  }
//  cout<<"this"<<endl;
  for(int i=1;i<=n;i++)cout<<b[i]<<' ';
  return 0;
}