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=505*505;
int n;
int a[N];
struct Node{
	int i,x;
	friend bool operator < (Node A,Node B){
		return A.x<B.x;
	}
}x[N];
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&x[i].x),x[i].i=i;
	sort(x+1,x+n+1);
	int L=1,R=n*n;
	memset(a,0,sizeof a);
	for (int i=1;i<=n;i++)
		a[x[i].x]=x[i].i;
	for (int i=1;i<=n;i++){
		int t=x[i].i-1;
		while (t--){
			while (L<=n*n&&a[L])
				L++;
			if (L>x[i].x)
				return puts("No"),0;
			a[L]=x[i].i;
		}
	}
	for (int i=n;i>=1;i--){
		int t=n-x[i].i;
		while (t--){
			while (R>=1&&a[R])
				R--;
			if (R<x[i].x)
				return puts("No"),0;
			a[R]=x[i].i;
		}
	}
	puts("Yes");
	for (int i=1;i<=n*n;i++)
		printf("%d ",a[i]);
	return 0;
}