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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
const int N=510;
struct data{int pos,w;}dt[N];
inline bool cmp(const data &a,const data &b){return a.pos<b.pos;}
int a[N*N];
int que[N],use[N];
int last[N],rk[N*N];
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&dt[i].pos);
		dt[i].w=i;a[dt[i].pos]=i;
	}
	sort(dt+1,dt+n+1,cmp);
	int sum=0,now=1;
	for(int i=1;i<=n;i++)
	{
		if(dt[i].pos-sum<dt[i].w)
		{
			printf("No\n");
			return 0;
		}
		for(int j=1;j<dt[i].w;j++)
		{
			while(a[now]) now++;
			a[now]=dt[i].w;
		}
		sum+=dt[i].w;
	}
	for(int i=1;i<=n;i++)
	for(int j=1;j<=n-dt[i].w;j++)
	{
		while(a[now]) now++;
		a[now]=dt[i].w;
	}
	for(int i=1;i<=n*n;i++)
	{
		rk[i]=rk[last[a[i]]]+1;
		last[a[i]]=i;
	}
	for(int i=1;i<=n;i++)
	if(rk[dt[i].pos]!=dt[i].w)
	{
		printf("No\n");
		return 0;
	}
	printf("Yes\n");
	for(int i=1;i<=n*n;i++) printf("%d ",a[i]);
	printf("\n");
	return 0;
}