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>
using namespace std;
#include<vector>
pair<int,int>p[500];
#include<algorithm>
int ans[500*500];
int main()
{
	int n;cin>>n;
	for(int i=0;i<n;p[i].second=i,i++)cin>>p[i].first;
	sort(p,p+n);
	int les=0,pre=0;
	for(int i=0;i<n;i++)
	{
		les+=p[i].first-pre-1;
		les-=p[i].second;
		if(les<0)
		{
			cout<<"No"<<endl;
			return 0;
		}
		int cnt=0;
		for(int j=0;cnt<p[i].second;)
		{
			if(ans[j]);
			else ans[j]=p[i].second+1,cnt++;
			j++;
			if(j>=n*n)
			{
				cout<<"No"<<endl;
				return 0;
			}
		}
		ans[p[i].first-1]=p[i].second+1;
		pre=p[i].first;
	}
	for(int i=1;i<=n;i++)
	{
		int cnt=n-i;
		int lastp=0;
		for(int j=0;j<n*n;j++)if(ans[j]==i)lastp=j;
		for(int j=lastp+1;cnt;)
		{
			if(ans[j]);
			else ans[j]=i,cnt--;
			j++;
			if(j>=n*n&&cnt>0)
			{
				cout<<"No"<<endl;
				return 0;
			}
		}
	}
	cout<<"Yes"<<endl;
	for(int i=0;i<n*n;i++)cout<<(i?" ":"")<<ans[i];
	cout<<endl;
}
