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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <memory.h>
#include <vector>

using namespace std;
typedef long long LL;
const int maxn = 251005;

int a[maxn],x[maxn],ord[maxn],pos,n;
bool comp(int u,int v) {return x[u]<x[v];}
int main()
{
	#ifdef Amberframe
		freopen("agc008d.in","r",stdin);
		freopen("agc008d.out","w",stdout);
	#endif
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
		scanf("%d",&x[i]),a[x[i]]=i,ord[i]=i;
	sort(ord+1,ord+n+1,comp);pos=1;
	for (int i=1;i<=n;i++) {
		int t=ord[i];
		for (int j=1;j<t;j++) {
			while (a[pos]) ++pos;
			a[pos]=t;++pos;
		}
		if (pos>x[t]) {puts("No");return 0;}
	}
	for (int i=1;i<=n;i++) {
		int t=ord[i];
		for (int j=n;j>t;j--) {
			while (a[pos]) ++pos;
			if (pos<x[t]) {puts("No");return 0;}
			a[pos]=t;++pos;
		}
	}
	puts("Yes");
	for (int i=1;i<=n*n;i++)
		printf("%d ",a[i]);
	return 0;
}