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;

int n,x[505],y[505],a[250010],cnt[505];

bool cmp(int a,int b){return x[a]<x[b];}

void work()
{
	scanf("%d",&n);
	for (int i=1; i<=n; i++)  scanf("%d",&x[i]),a[x[i]]=i,cnt[i]=i-1,y[i]=i;
	sort(y+1,y+n+1,cmp);
	for (int i=1,j=1; i<=n*n; i++)
		if (!a[i])
			{
				while ((j<=n)&&(!cnt[y[j]]))  j++;
				if (j<=n)  a[i]=y[j],cnt[y[j]]--;
			}else  if (cnt[a[i]])  puts("No"),exit(0);
	for (int i=1; i<=n; i++)  cnt[i]=n-i;
	for (int i=1,j=1; i<=n*n; i++)
		if (!a[i])
			{
				while ((j<=n)&&(!cnt[y[j]]))  j++;
				if (x[y[j]]<i)  a[i]=y[j],cnt[y[j]]--;
				else  puts("No"),exit(0);
			}
	puts("Yes");
	for (int i=1; i<=n*n; i++)  printf("%d ",a[i]);	
}

int main()
{
	work();
	return 0;
}
