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;
#define MAXN 510
#define MAXM 250010
int a[MAXM],f[MAXN],x[MAXN],n,ls=1;
bool cmp(int a,int b)
{
	return x[a]<x[b];
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&x[i]);
		a[x[i]]=i;
		f[i]=i;
	}
	sort(f+1,f+1+n,cmp);
	for(int i=1;i<=n;i++)
	{
		int k=f[i];
		for(int j=1;j<k;j++)
		{
			while(a[ls]) ls++;
			if(ls>x[k])
			{
				printf("No\n");
				return 0;
			}
			a[ls]=k;
		}
	}
	for(int i=1;i<=n;i++)
	{
		int k=f[i];
		for(int j=1;j<=n-k;j++)
		{
			while(a[ls]) ls++;
			if(ls<x[k]||ls>n*n)
			{
				printf("No\n");
				return 0;
			}
			a[ls]=k;
		}
	}
	printf("Yes\n");
	for(int i=1;i<=n*n;i++)
		printf("%d ",a[i]);
}