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=3e5+5;
int n,a[N];
struct each{
	int a,b;
}b[N];
bool com(each x,each y){
	return x.a<y.a;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&b[i].a);
		b[i].b=i,a[b[i].a]=i;
	}
	int now=1;
	sort(b+1,b+n+1,com);
	for(int i=1;i<=n;i++){
		for(int j=1;j<b[i].b;j++){
			while(now<=n*n&&a[now])	++now;
			if(now>b[i].a){
				puts("No");
				return 0;
			}
			a[now++]=b[i].b;
		}
	}

	now=n*n;
	for(int i=n;i;i--){
		for(int j=1;j<=n-b[i].b;j++){
			while(now<=n*n&&a[now])	--now;
			if(now<b[i].a){
				puts("No");
				return 0;
			}
			a[now--]=b[i].b;
		}
	}

	puts("Yes");
	for(int i=1;i<=n*n;i++)	printf("%d ",a[i]);
	return 0;
}