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 <bits/stdc++.h>
using namespace std;
pair<int,int> x[505];
int ans[250005],c[505];
int main()
{
	int n,cur=1;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&x[i].first);
		x[i].second=i;
		ans[x[i].first]=i;
	}
	sort(x+1,x+n+1);
	for (int i=1;i<=n;i++)
	{
		int cnt=x[i].second-1;
		while (cnt--)
		{
			while (ans[cur])
			cur++;
			ans[cur]=x[i].second;
		}
	}
	for (int i=1;i<=n;i++)
	{
		int cnt=n-x[i].second;
		while (cnt--)
		{
			while (ans[cur])
			cur++;
			ans[cur]=x[i].second;
		}
	}
	for (int i=1;i<=n*n;i++)
	{
		c[ans[i]]++;
		if (c[ans[i]]==ans[i])
		{
			for (int j=1;j<=n;j++)
			{
				if (x[j].second==ans[i] && x[j].first!=i)
				{
					printf("No");
					return 0;
				}
			}
		}
	}
	printf("Yes\n");
	for (int i=1;i<=n*n;i++)
	printf("%d ",ans[i]);
}