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>
#define P pair<int,int>
using namespace std;

vector<int>ans;
P c[500];
signed main(){
	int a;cin>>a;
	for(int b=0;b<a;b++){
		int d;cin>>d;d--;
		c[b]={d,b};
		}	
	sort(c,c+a);
	int count[500]{};
	for(int b=0;b<a*a;b++){
		bool S=false;
		for(int d=0;d<a;d++){
			if(c[d].first==b){
				if(count[c[d].second]==c[d].second){
					ans.push_back(c[d].second+1);
					count[c[d].second]++;
					S=true;
					}
				else {puts("No");return 0;}	
				}
			}
		if(S)continue;	
		for(int d=0;d<a;d++){
				if(count[c[d].second]<c[d].second){
					ans.push_back(c[d].second+1);
					count[c[d].second]++;
					S=true;
					break;
					}
			}
		if(S)continue;
			for(int d=0;d<a;d++){
				if(c[d].first<b&&count[c[d].second]<a){
					ans.push_back(c[d].second+1);
					count[c[d].second]++;
					S=true;
					break;
					}
			}		
		}	
		if(ans.size()!=a*a){puts("No");return 0;}
		puts("Yes");
		for(int b=0;b<a*a;b++){
		if(b)cout<<" ";
		cout<<ans[b];
		}	
		cout<<endl;
	}
