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;
#define RI register int
int a[250005],x[505],n;
struct node{int v,x;}t[505];
bool cmp(node k1,node k2) {return k1.x<k2.x;}
int main()
{
	int x;scanf("%d",&n);
	for(RI i=1;i<=n;++i) scanf("%d",&x),a[x]=i,t[i]=(node){i,x};
	sort(t+1,t+1+n,cmp);
	int j=1;
	for(RI i=1;i<=n;++i) {
		for(RI k=1;k<=t[i].v-1;++k) {
			while(j<n*n&&a[j]) ++j;
			if(a[j]||j>t[i].x) {puts("No");return 0;}
			a[j]=t[i].v;
		}
	}
	for(RI i=1;i<=n;++i) {
		for(RI k=t[i].v+1;k<=n;++k) {
			while(j<n*n&&a[j]) ++j;
			if(a[j]||j<t[i].x) {puts("No");return 0;}
			a[j]=t[i].v;
		}
	}
	puts("Yes");
	for(RI i=1;i<=n*n;++i) printf("%d ",a[i]);
	return 0;
}