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;

const int N=600;
pair<int,int>a[N];
int b[N*N],n;
void gofail(){cout<<"No\n";exit(0);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;++i)cin>>a[i].first,a[i].second=i,b[a[i].first]=i;
	sort(a+1,a+n+1);
	for(int i=1;i<=n;++i){
		int les=a[i].second-1;
		for(int j=1;j<a[i].first&&les;++j)
			if(!b[j])b[j]=a[i].second,--les;
		if(les)gofail();
	}
	for(int i=n;i;--i){
		int les=n-a[i].second;
		for(int j=n*n;j>a[i].first&&les;--j)
			if(!b[j])b[j]=a[i].second,--les;
		if(les)gofail();
	}
	cout<<"Yes\n";
	for(int i=1;i<=n*n;++i)
		cout<<b[i]<<" \n"[i==n*n];
	return 0;
}
