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;
typedef long long int uli;
const int mx=512;
int d[mx];
int o[mx];
int ans[mx*mx];
bool cmp(int i,int j){ return d[i]<d[j]; }
int main(){
  int n;
  scanf("%d",&n);
  for(int i=1;i<=n;i++){
    scanf("%d",d+i);
    o[i]=i;
    ans[d[i]]=i;
  }
  sort(o+1,o+1+n,cmp);
  int lft=1;
  bool ok=true;
  for(int i=1;i<=n && ok;i++){
    int v=o[i];
    int cnt=0;
    while(lft<d[v] && cnt<v-1){
      if(ans[lft]==0){
        ans[lft]=v;
        cnt++;
      }
      lft++;
    }
    if(cnt<v-1)ok=false;
  }
  int rht=n*n;
  for(int i=n;i>=1;i--){
    int v=o[i];
    int cnt=0;
    while(rht>d[v] && cnt<n-v){
      if(ans[rht]==0){
        ans[rht]=v;
        cnt++;
      }
      rht--;
    }
    if(cnt<n-v)ok=false;
  }
  if(ok){
    puts("Yes");
    for(int i=1;i<=n*n;i++){
      if(i!=1)printf(" ");
      printf("%d",ans[i]);
    }
    puts("");
  }
  else puts("No");
  return 0;
}
