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 <algorithm>
#include <queue>
#include <cassert>
using namespace std; typedef long long ll; const int MINF=-1e9; typedef pair<int,int> P;

int main() {
  int n; cin>>n;
  int x[n]; for(int i=0;i<n;i++) cin>>x[i];
  int a[n*n]; fill(a,a+n*n,0);
  for(int i=0;i<n;i++) {
    --x[i];
    if (a[x[i]] != 0) return puts("No")*0;
    a[x[i]]=i+1;
  }

  P p[n];
  for(int i=0;i<n;i++) {
    p[i]={x[i],i}; // i-1個i+1が必要
  }
  sort(p,p+n);
  int j=0;
  for(P y:p) {
    for(int i=0;i<y.second;i++) {
      while(a[j]!=0) {
        j++; if (j>=y.first) return puts("No")*0; // j<n*nも伴う
      }
      a[j]=y.second+1;
    }
  }
  // この時点で、o[i]=i+1（残りn-1-i）
  j=0;
  for(P y:p) {
    for(int i=0;i<n-1-y.second;i++) {
      while(a[j]!=0 || j<y.first) {
        j++; if (j>=n*n) return puts("No")*0; // j<n*nも伴う
      }
      a[j]=y.second+1;
    }
  }

  cout<<"Yes"<<endl;
  for(int i=0;i<n*n;i++) cout<<a[i]<<" ";
  cout<<endl;
}