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;
typedef long long int ll;
#define FOR(i,l,r) for(i=l;i<r;i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(x) x.begin(),x.end()
#define P pair<ll,ll>
#define F first
#define S second
signed main(){
  ll N,i,j=0;cin>>N;ll A[N],B[N],X[N],Y[N];
  REP(i,N)cin>>A[i];REP(i,N)cin>>B[i];
  REP(i,N){
    X[i]=upper_bound(A,A+N,i+1)-lower_bound(A,A+N,i+1);
    Y[i]=upper_bound(B,B+N,i+1)-lower_bound(B,B+N,i+1);
    if(X[j]+Y[j]<X[i]+Y[i])j=i;
  }
  if(X[j]+Y[j]>N)cout<<"No"<<endl;
  else{
    cout<<"Yes"<<endl;ll x=0;
    while(1){
      REP(i,N)if(A[i]==B[(i+x)%N])break;
      if(i==N)break;x++;
    }
    REP(i,N-1)cout<<B[(i+x)%N]<<" ";
    cout<<B[(N-1+x)%N]<<endl;
  }
  return 0;
}