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<cmath>
#include<stdio.h>
#include<algorithm>
#define ll long long
using namespace std;
const int N=1000100;
inline ll read(){
	ll x=0,f=0,c=getchar();
	for(;c>'9'||c<'0';f=c=='-',c=getchar());
	for(;c>='0'&&c<='9';c=getchar())
	x=(x<<1)+(x<<3)+c-'0';return f?-x:x;
}
struct cqz{int x,i;};
inline bool operator<(cqz i,cqz j){
	return i.x>j.x;
}
#include<queue>
priority_queue<cqz>q;
int n,i,j,a[N],x[N];
int main(){
	//freopen("1.out","w",stdout);
	n=read();
	for(i=1;i<=n;i++){
		x[i]=read();
		if(a[x[i]])goto fuck;
		a[x[i]]=i;
		for(j=1;j<i;j++)
		q.push((cqz){x[i],i});
		x[i]=i;
	}
	for(i=1;i<=n*n;i++)
	if(a[i]){
		x[a[i]]--;
		if(x[a[i]])goto fuck;
		for(j=a[i]+1;j<=n;j++)
		q.push((cqz){n*n+1,a[i]});
	}else{
		if(q.empty())goto fuck;
		a[i]=q.top().i;q.pop();
		x[a[i]]--;
	}
	puts("Yes");
	for(i=1;i<=n*n;i++)printf("%d ",a[i]);
	return 0;fuck:puts("No");
}