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 ll;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

int A[505*505];

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  
  int N; cin>>N;
  vector<pair<int,int> > X(N);
  for(int i=1;i<=N;i++){
    int x; cin>>x;
    X[i-1]={x,i};
    A[x]=i;
  }
  sort(X.begin(),X.end());
  int l=1;
  for(auto p:X){
    int x=p.first, num=p.second;
    int cnt=num-1;
    while(l<x&&cnt>0){
      if(A[l]==0){
        A[l]=num;
        cnt--;
      }
      l++;
    }
    if(cnt>0){
      cout<<"No\n";
      return 0;
    }
  }
  reverse(X.begin(),X.end());
  int r=N*N;
  for(auto p:X){
    int x=p.first, num=p.second;
    int cnt=N-num;
    while(r>x&&cnt>0){
      if(A[r]==0){
        A[r]=num;
        cnt--;
      }
      r--;
    }
    if(cnt>0){
      cout<<"No\n";
      return 0;
    }
  }
  cout<<"Yes\n";
  for(int i=1;i<=N*N;i++) cout<<A[i]<<" ";
  cout<<"\n";
  return 0;
}
