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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=505;
struct data{
	int x,id,now;
}a[N],b[N];
bool cmp(data a,data b){
	return a.x<b.x;
}
int n,now,x[N*N];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i].x);
		if(a[i].x<i){
			puts("No");
			return 0;
		}
		a[i].id=i;
	}
	sort(a+1,a+n+1,cmp);
	for(int i=1;i<=n;i++){
		while(a[i].now<a[i].id-1){
			now++;
			if(x[now]){
				continue;
			}
			x[now]=a[i].id;
			a[i].now++;
		}
		if(now>=a[i].x){
			puts("No");
			return 0;
		}
		x[a[i].x]=a[i].id;
		a[i].now++;
	}
	now=n*n+1;
	for(int i=n;i>=1;i--){
		while(a[i].now<n){
			now--;
			if(x[now]){
				continue;
			}
			x[now]=a[i].id;
			a[i].now++;
		}
		if(now<=a[i].x){
			puts("No");
			return 0;
		}
		x[a[i].x]=a[i].id;
		a[i].now++;
	}
	puts("Yes");
	for(int i=1;i<=n*n;i++){
		printf("%d ",x[i]);
	}
	puts("");
	return 0;
}