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<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N=3e5;
struct Node{int w,id;bool operator <(const Node &a)const{return w>a.w;}};
priority_queue<Node>q;
int p[N],x[N],a[N],cnt[N],n,tmp[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",x+i),p[x[i]]=i;
	for(int i=2;i<=n;i++)q.push((Node){x[i],i});
	int ti=0,tn=0;
	for(int i=1;i<=n*n;i++)
		if(p[i])
		{
			if(cnt[p[i]]!=p[i]-1){puts("No");return 0;}
			a[i]=p[i];for(int j=1;j<=n-p[i];j++)tmp[++tn]=p[i];
		}
		else
		{
			if(!q.empty())
			{
				Node t=q.top();q.pop();
				a[i]=t.id;cnt[t.id]++;
				if(cnt[t.id]!=t.id-1)q.push((Node){t.w,t.id});
			}
			else if(ti<tn)a[i]=tmp[++ti];
			else{puts("No");return 0;}
		}
	puts("Yes");
	for(int i=1;i<=n*n;i++)printf("%d ",a[i]);
}