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<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int x[600],a[600],n,A[300000];
int compare(int k1,int k2){
	return x[k1]<x[k2];
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&x[i]),a[i]=i;
	sort(a+1,a+n+1,compare);
	for (int i=1;i<n;i++) if (x[a[i]]==x[a[i+1]]){
		printf("No\n"); return 0;
	}
	int now=1;
	for (int i=1;i<=n;i++){
		for (int j=1;j<a[i];j++){
			while (now<x[a[i]]&&A[now]) now++;
			A[now]=a[i]; now++;
		}
	//	cout<<i<<" "<<now<<endl;
		if (now>x[a[i]]){
			printf("No\n"); return 0;
		}
		A[x[a[i]]]=a[i];
	}
	now=1;
	for (int i=1;i<=n;i++){
		while (A[now]) now++;
		if (now<x[a[i]]){
			printf("No\n"); return 0;
		}
		for (int j=1;j<=n-a[i];j++){
			while (A[now]) now++;
			A[now]=a[i]; now++;
		}
	}
	printf("Yes\n");
	for (int i=1;i<=n*n;i++) printf("%d ",A[i]);
	return 0;
}
		