You are given an integer sequence x of length N. Determine if there exists an integer sequence a that satisfies all of the following conditions, and if it exists, construct an instance of a.

* a is N^2 in length, containing N copies of each of the integers 1, 2, ..., N.
* For each 1 ≤ i ≤ N, the i-th occurrence of the integer i from the left in a is the x_i-th element of a from the left.

Constraints

* 1 ≤ N ≤ 500
* 1 ≤ x_i ≤ N^2
* All x_i are distinct.

Input

The input is given from Standard Input in the following format:


N
x_1 x_2 ... x_N


Output

If there does not exist an integer sequence a that satisfies all the conditions, print `No`. If there does exist such an sequence a, print `Yes` in the first line, then print an instance of a in the second line, with spaces inbetween.

Examples

Input

3
1 5 9


Output

Yes
1 1 1 2 2 2 3 3 3


Input

2
4 1


Output

No
#include<bits/stdc++.h>
using namespace std;
int main(){
  int N;cin>>N;
  vector<pair<int,int> >X(N);
  for(int i=0;i<N;i++){cin>>X[i].first;X[i].second=i+1;}
  sort(X.begin(),X.end());

  vector<int>A(N*N,-1);
  //セグ木で高速化?
  int l=0;
  for(int i=0;i<N;i++){
    A[X[i].first-1]=X[i].second;
    for(int j=0;j<X[i].second-1;j++){
      while(A[l]!=-1)l++;
      A[l]=X[i].second;
    }
    if(l>X[i].first-1){cout<<"No";return 0;}
  }
  for(int i=0;i<N;i++){
    for(int j=0;j<N-X[i].second;j++){
      while(A[l]!=-1)l++;
      if(l<X[i].first-1){cout<<"No";return 0;}
      A[l]=X[i].second;
    }
  }
  cout<<"Yes"<<endl;
  for(int i=0;i<N*N;i++)cout<<A[i]<<" ";

  

}