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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)

using namespace std;

const int N=510;
int n,b[N*N];
struct node{int x,i;}a[N];

const bool cmp(node x,node y){return x.x<y.x;}

int main(){
	scanf("%d",&n);
	rep (i,1,n) scanf("%d",&a[i].x),a[i].i=i;
	sort(a+1,a+1+n,cmp);
	rep (i,1,n){
		int tmp=a[i].i-1;
		rep (j,1,a[i].x-1){
			if (!tmp) break;
			if (!b[j]) b[j]=a[i].i,tmp--;
		}
		if (tmp) return puts("No"),0;
		b[a[i].x]=a[i].i;
	}
	for (int i=n;i;i--){
		int tmp=n-a[i].i;
		rep (j,a[i].x+1,n*n){
			if (!tmp) break;
			if (!b[j]) b[j]=a[i].i,tmp--;
		}
		if (tmp) return puts("No"),0;
	}
	puts("Yes");
	rep (i,1,n*n) printf("%d ",b[i]);
	return 0;
}