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<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;

const int N=505;

int n,x[N],f[N],a[N*N+N];

bool cmp(int p,int q)
{
    return x[p]<x[q];
}

int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++) scanf("%d",&x[i]),f[i]=i;
    sort(f+1,f+n+1,cmp);
    int now=1;
    for (int i=1;i<=n;i++)
    {
        int w=f[i];a[x[w]]=w;
        for (int j=1;j<w;j++)
        {
            while (a[now]) now++;
            a[now]=w;
        }
        if (now>x[w]) {puts("No");return 0;}
    }
    for (int i=1;i<=n;i++)
    {
        int w=f[i];
        for (int j=1;j<=n-w;j++)
        {
            while (a[now]) now++;
            if (now<x[w]) {puts("No");return 0;}
            a[now]=w;
        }
    }
    puts("Yes");
    for (int i=1;i<=n*n;i++) printf("%d ",a[i]);
    return 0;
}