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;
  scanf("%d",&n);
  int x[500];
  for(int i=0;i<n;i++){
    scanf("%d",x+i);
    x[i]--;
  }
  priority_queue<pair<int,int> > Q;
  for(int i=0;i<n;i++){
    for(int k=0;k<i;k++){
      Q.push(make_pair(-x[i],i));
    }
  }
  static int A[500000];
  for(int i=0;i<n*n;i++){
    A[i]=-1;
  }
  for(int i=0;i<n;i++){
    A[x[i]]=i;
  }
  int T[500];
  for(int i=0;i<n;i++){
    T[i]=i;
  }
  bool F=0;
  for(int i=0;i<n*n;i++){
    if(A[i]!=-1){
      int j=A[i];
      if(T[j]!=0){
	F=1;
	break;
      }
      for(int k=0;k<n-j-1;k++){
	Q.push(make_pair(-n*n-1,j));
      }
    }
    else{
      if(Q.empty()){
	F=1;
	break;
      }
      A[i]=Q.top().second;
      T[A[i]]--;
      Q.pop();
    }
  }
  if(F){
    puts("No ");
  }
  else{
    puts("Yes ");
    for(int i=0;i<n*n;i++){
      printf("%d ",A[i]+1);
    }
    putchar('\n');
  }
  return 0;
}
