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 <cstdio>
#include <cstring>
#include <algorithm>
#include <utility>
#define FR first
#define SE second

using namespace std;

typedef pair<int,int> pr;

int num[250005];
int ans[250005];
pr a[505];

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++) {
  	int x;
  	scanf("%d",&x);
  	a[i]=pr(x,i);
  	ans[x]=i;
  }
  sort(a+1,a+n+1);
  for(int i=1;i<=n;i++) {
  	int x=a[i].SE,tot=0;
  	for(int j=1;j<=n*n;j++)
  	  if (!ans[j]) num[++tot]=j;
  	int p=lower_bound(num+1,num+tot+1,a[i].FR)-num-1;
  	if (p<x-1) {
  		puts("No");
  		return 0;
	  }
	for(int j=1;j<x;j++) ans[num[j]]=x;
  }
  for(int i=n;i>0;i--) {
  	int x=a[i].SE,tot=0;
  	for(int j=1;j<=n*n;j++)
  	  if (!ans[j]) num[++tot]=j;
  	int p=lower_bound(num+1,num+tot+1,a[i].FR)-num;
  	if (tot-p+1<n-x) {
  		puts("No");
  		return 0;
	  }
	for(int j=1;j<=n-x;j++) ans[num[tot-j+1]]=x;
  }
  puts("Yes");
  for(int i=1;i<=n*n;i++) printf("%d ",ans[i]);
  printf("\n");
  return 0;
}