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 <cstdio>
#include <algorithm>
using namespace std;
const int maxn=510;
#define fi first
#define se second
#define mp make_pair

int n;
pair<int,int> x[maxn];
int a[maxn*maxn];

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&x[i].fi);
		x[i].se=i;
	}
	
	sort(x+1,x+n+1);
	int s=0;
	for(int i=1;i<=n;i++)
	{
		s+=x[i].se;
		if (s>x[i].fi)
		{
			printf("No\n");
			return 0;
		}
	}
	
	s=0;
	for(int i=n;i>=1;i--)
	{
		s+=n-x[i].se+1;
		if (s>n*n-x[i].fi+1)
		{
			printf("No\n");
			return 0;
		}
	}
	
	printf("Yes\n");
	for(int i=1;i<=n;i++)
	{
		int pos=1;
		a[x[i].fi]=x[i].se;
		for(int j=1;j<x[i].se;j++)
		{
			while(a[pos]!=0) pos++;
			a[pos]=x[i].se;
	    }
	}
	
	for(int i=1;i<=n;i++)
	{
		int pos=x[i].fi;
		for(int j=x[i].se+1;j<=n;j++)
		{
			while(a[pos]!=0) pos++;
			a[pos]=x[i].se;
		}
	}
	
	for(int i=1;i<=n*n;i++)
		printf("%d ",a[i]);
	printf("\n");
	return 0;
}