Problem:
Score : 800 points

Problem Statement

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.

Sample Input 1

3
1 5 9

Sample Output 1

Yes
1 1 1 2 2 2 3 3 3

For example, the second occurrence of the integer 2 from the left in a in the output is the fifth element of a from the left.
Similarly, the condition is satisfied for the integers 1 and 3.

Sample Input 2

2
4 1

Sample Output 2

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int main(){
  int n,p;
  cin>>n;
  int x[n],a[n*n],u[n];
  vector<P> v;
  for(int i=0;i<n*n;i++) a[i]=0;
  for(int i=0;i<n;i++){
    cin>>x[i];
    u[i]=1;
    a[x[i]-1]=i+1;
    v.push_back(P(x[i]-1,i+1));
  }
  sort(v.begin(),v.end());
  bool f=1;
  p=0;
  for(int i=0;i<n;i++){
    int c=v[i].second;
    while(u[c-1]<c){
      if(a[p]){
	p++;
	continue;
      }
      if(p>v[i].first){
	f=0;
	break;
      }
      a[p]=c;
      u[c-1]++;
    }
  }
  reverse(v.begin(),v.end());
  p=n*n-1;
  for(int i=0;i<n;i++){
    int c=v[i].second;
    while(u[c-1]<n){
      while(a[p]) p--;
      if(p<v[i].first){
	f=0;
	break;
      }
      a[p]=c;
      u[c-1]++;
    }
  }

  //assertion
  if(f){
    for(int i=1;i<=n;i++){
      int t=0;
      for(int j=0;j<n*n;j++){
	if(a[j]==i) t++;
	if(i==t) {
	  f&=(j+1==x[i-1]);
	  break;
	}
      }
    }
  }
  if(f){
    for(int i=1;i<=n;i++){
      int t=0;
      for(int j=0;j<n*n;j++){
	if(a[j]==i) t++;
      }
      //cout<<i<<" "<<t<<endl;
      f&=(t==n);
    }
  }
  if(!f){
    cout<<"No"<<endl;
  }else{
    cout<<"Yes"<<endl;
    for(int i=0;i<n*n;i++) cout<<a[i]<<" \n"[i==n*n-1];
    //for(int i=0;i<n*n;i++) cout<<i<<":"<<a[i]<<endl;
  }
  return 0;
}