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 <algorithm>
using namespace std;
int n,nn,i,j,c,r[505*505];
pair<int,int> a[505];
int main() {
  scanf("%d",&n); nn=n*n;
  for (i=0; i<n; i++) {
    scanf("%d",&a[i].first);
    a[i].second=i+1;
  }
  sort(a,a+n);
  for (i=0, j=1; i<n; i++) {
    for (c=1; j<=nn && c<a[i].second; j++) if (r[j]==0) {
      r[j]=a[i].second;
      c++;
    }
    if (j>nn || c<a[i].second || r[a[i].first]!=0) { puts("No"); return 0; }
    r[a[i].first]=a[i].second;
  }
  reverse(a,a+n);
  for (i=0, j=nn; i<n; i++) {
    for (c=0; j>0 && c<n-a[i].second; j--) if (r[j]==0) {
      r[j]=a[i].second;
      c++;
    }
    if (j<a[i].first || c<n-a[i].second) { puts("No"); return 0; }
  }
  puts("Yes");
  for (i=1; i<=nn; i++) printf("%d%c",r[i],(i==nn)?'\n':' ');
  return 0;
}
