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<iostream>
#include<algorithm>
#include<string>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<stack>
#include<utility>
#include<iomanip>
using namespace std;

int main()
{
	int n;cin>>n;
	int a[501];
	for(int i=1;i<=n;i++)cin>>a[i];
	int s[250010];
	memset(s,-1,sizeof(s));
	int l=0,r=n*n-1;
	for(int i=1;i<=n;i++)
	{
		s[a[i]-1]=i;
	}
	for(int i=1;i<=n;i++)
	{
		for(int c=1;c<i&&l<n*n;)
		{
			while(s[l]!=-1&&l<n*n)l++;
			if(l==n*n)
			{
				cout<<"No"<<endl;
				return 0;
			}
			s[l++]=i;
			c++;
		}
	}
	for(int i=n;i>0;i--)
	{
		for(int c=0;c<n-i&&r>=0;)
		{
			while(s[r]!=-1&&r>=0)r--;
			if(r<0)
			{
				cout<<"No"<<endl;
				return 0;
			}
			s[r--]=i;
			c++;
		}
	}
	bool flag=true;
	int p[501][501];
	memset(p,-1,sizeof(p));
	int cou[501];
	memset(cou,0,sizeof(cou));
	for(int i=0;i<n*n;i++)
	{
		if(s[i]!=-1)p[s[i]][cou[s[i]]++]=i;
	}
	for(int i=1;i<=n;i++)
	{
		if(p[i][i-1]!=a[i]-1)flag=false;
		if(p[i][n-1]==-1)flag=false;
	}
	if(flag)
	{
		cout<<"Yes"<<endl;
		for(int i=0;i<n*n;i++)cout<<s[i]<<(i==n*n-1?"\n":" ");
	}
	else cout<<"No"<<endl;
}