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<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
struct X
{
	int pos,i;
	friend bool operator < (X a,X b){return a.pos<b.pos;}
} x[505];
int ans[250005];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&x[i].pos),x[i].i=i;
	sort(x+1,x+n+1);
	int p=1;
	for(int i=1;i<=n;i++)
	{
		ans[x[i].pos]=x[i].i;
		for(int j=1;j<=x[i].i-1;j++)
		{
			while(ans[p]&&p<x[i].pos) p++;
			if(p==x[i].pos){printf("No");return 0;}
			ans[p++]=x[i].i;
		}
	}
	for(int i=1;i<=n;i++)
	{
		if(p<=x[i].pos-1&&!ans[p]){printf("No");return 0;}
		for(int j=1;j<=n-x[i].i;j++)
		{
			while(ans[p]) p++;
			ans[p++]=x[i].i;
		}
		//if(i<n&&p<x[i+1].pos-1){printf("No");return 0;}
	}
	printf("Yes\n");
	for(int i=1;i<=n*n;i++) printf("%d ",ans[i]);
	return 0;
}