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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define all(v) v.begin(),v.end()
#define P pair<int,int>
#define len(s) (int)s.size()
#define pb push_back

template<class T> inline bool chmin(T &a, T b){
	if(a>b){a=b;return true;}
	return false;
}
template<class T> inline bool chmax(T &a, T b){
	if(a<b){a=b;return true;}
	return false;
}
constexpr int mod = 1e9+7;
constexpr int inf = 3e18;

int N;
P X[505];
int ans[500*505];
signed main(){
	cin>>N;
	REP(i,N+1){
		cin>>X[i].first;X[i].second=i;
		ans[X[i].first]=i;
	}
	sort(X+1,X+N+1);
	for(int i=N;i>0;i--){
		int j=X[i].second-1,k=X[i].first-1;
		while(j&&k){
			if(!ans[k]){
				ans[k]=X[i].second;j--;
			}
			k--;
		}
		if(j){
			cout<<"No"<<endl;return 0;
		}
	}
	REP(i,N+1){
		int j=N-X[i].second,k=X[i].first+1;
		while(j&&k<=N*N){
			if(!ans[k]){
				ans[k]=X[i].second;j--;
			}
			k++;
		}
		if(j){
			cout<<"No"<<endl;return 0;
		}
	}
	cout<<"Yes"<<endl;
	REP(i,N*N+1)cout<<ans[i]<<" ";
	cout<<endl;
}