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;
using ll=long long;
using P=pair<ll,ll>;
ll N;
ll a[300000];
int main(){
  cin>>N;
  for(ll i=1;i<=N*N;i++)
    a[i]=-1;
  vector<P>A(N);
  for(ll i=0;i<N;i++){
  ll x;cin>>x;P p(x,i+1);
    A[i]=p;
   }bool judge=true;sort(A.begin(),A.end());
  ll d=1;
  for(ll i=0;i<N;i++){
  ll x=A[i].first;ll n=A[i].second;
    a[x]=n;
    ll sum=1;
    while(sum<n){
    if(a[d]==-1){a[d]=n;sum++;d++;}
      else
        d++;
    if(x<d)judge=false;
    }if(!judge)break;
  }if(!judge)
    cout<<"No"<<endl;
  else{
    ll d1=N*N;
  for(ll i=N-1;0<=i;i--){
  ll x=A[i].first;ll n=A[i].second;
    ll sum=0;
    while(sum<N-n){
    if(a[d1]==-1){a[d1]=n;sum++;d1--;}
      else
        d1--;
    if(d1<x)judge=false;
    }if(!judge)break;
  }if(!judge)
    cout<<"No"<<endl;
    else{
      cout<<"Yes"<<endl;
    for(ll i=1;i<N*N;i++)
      cout<<a[i]<<" ";
    cout<<a[N*N]<<endl;
    }
  
  }
  return 0;
}