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<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>

#define maxn 505

using namespace std;

inline int getint()
{
	int num=0,flag=1;char c;
	while((c=getchar())<'0'||c>'9')if(c=='-')flag=-1;
	while(c>='0'&&c<='9')num=num*10+c-48,c=getchar();
	return num*flag;
}

int n;
int p[maxn],f[maxn],a[maxn*maxn];
inline bool cmp(int x,int y){return p[x]<p[y];}

int main()
{
	while(~scanf("%d",&n))
	{
		bool flag=0;
		for(int i=1;i<=n;i++)p[i]=getint(),f[i]=i;
		sort(f+1,f+n+1,cmp);
		int now=1;
		for(int i=1;i<=n;i++)
		{
			a[p[f[i]]]=f[i];
			for(int j=1;j<f[i];j++)
			{
				while(a[now])now++;
				a[now]=f[i];
			}
			if(now>p[f[i]]){puts("No");flag=1;break;}
		}
		if(flag)continue;
		for(int i=1;i<=n;i++)
		{
			for(int j=1;j<=n-f[i];j++)
			{
				while(a[now])now++;
				if(now<p[f[i]]){puts("No");flag=1;break;}
				a[now]=f[i];
			}
			if(flag)break;
		}
		if(!flag)
		{
			puts("Yes");
			for(int i=1;i<=n*n;i++)printf("%d%c",a[i],i==n*n?10:32);
		}
	}
}