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>
#define M 505
#define XFZSB {printf("No\n"); return 0;}
#define YXQAK {printf("Yes\n");}
using namespace std;

struct node{
	int id,x; node(){id=x=0;}
	node(int iid,int nnum){id=iid; x=nnum;}
	friend bool operator <(node a,node b){return a.x<b.x;}
}a[M];
int num[M*M]={0},now=1; short sum[M*M][M]={0};
int main(){
	int n; scanf("%d",&n);
	for(int i=1;i<=n;i++){
		int x; scanf("%d",&x);
		a[i]=node(i,x);
	}
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++){
		num[a[i].x]=a[i].id; 
		int ci=sum[now][a[i].id]+1;
		while(now<=a[i].x&&ci<a[i].id){
			if(!num[now]){
				num[now]=a[i].id; ci++;
			}
			for(int j=1;j<=n;j++)
			sum[now][j]=sum[now-1][j]+(num[now]==j);
			now++;
		}
		//if(sum[a[i].x][a[i].id]!=a[i].id) XFZSB
	}
	for(int i=1;i<=n;i++){
		int ci=sum[now-1][a[i].id]+(now<=a[i].x);
		while(now<=n*n&&ci<n){
			if(!num[now]){
				num[now]=a[i].id; ci++;
			}
			for(int j=1;j<=n;j++)
			sum[now][j]=sum[now-1][j]+(num[now]==j);
			now++;
		}
	}
	for(int j=1;j<=n;j++)
	sum[n*n][j]=sum[n*n-1][j]+(num[now]==j);
	for(int i=1;i<=n;i++)
	if(sum[a[i].x][a[i].id]!=a[i].id) XFZSB;
	YXQAK
	for(int i=1;i<=n*n;i++) printf("%d ",num[i]);
}