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=505;
int ans[N*N],x[N],f[N],n;
int cmp(int a,int b){
	return x[a]<x[b];
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)scanf("%d",&x[i]),f[i]=i;
	sort(f+1,f+n+1,cmp);
	int k=1;
	for (int i=1;i<=n;i++){
		ans[x[f[i]]]=f[i];
		for (int j=1;j<f[i];j++){
			while (ans[k])k++;
			if (k>x[f[i]]){
				puts("No");
				return 0;
			}
			ans[k]=f[i];
		}
	}
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n-f[i];j++){
			while (ans[k])k++;
			if (k<x[f[i]]){
				puts("No");
				return 0;
			}
			ans[k]=f[i];
		}
	puts("Yes");
	for (int i=1;i<=n*n;i++)printf("%d ",ans[i]);
}