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 ll long long
#define rep(i,n) for(int i=0;i<n;i++)
const ll MOD=1000000007;
const long double PI=3.14159265358979;
const ll MAX=2000;


int main() {
  ll N,c=1;
  cin>>N;
  ll a[N];
  ll b[2*N];
  ll ac[N+1];
  ll ad[N+1];
  ll bd[N+1];
  ll bc[N+1];
  rep(i,N+1){
    ac[i]=0;bc[i]=0;
  }
  rep(i,N){
    cin>>a[i];
    ac[a[i]]++;
    ad[a[i]]=1;
  }
  rep(i,N){
    cin>>b[i];
    b[i+N]=b[i];
    bc[b[i]]++;
    bd[b[i]]=1;
  }
  ll x=0;
  rep(i,N+1){
    if(ac[i]+bc[i]>N){
      c=0;
    }
    if(i){
      ac[i]+=ac[i-1];
      bc[i]+=bc[i-1];
    }
  }
  for(int i=1;i<N+1;i++){
    if(ad[i]==1&&bd[i]==1){
      x=max(x,ac[i]-bc[i-1]);
     // cout<<ac[i]<<" "<<bc[i-1]<<"/";
    }
  }
  if(c){
    cout<<"Yes"<<endl;
    for(int i=N-x;i<2*N-x;i++){
      cout<<b[i]<<" ";
    }
  }
  else{
    cout<<"No";
  }
}

