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<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define MN 500
#define pa pair<int,int>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,a[MN*MN+5];
priority_queue<pa,vector<pa>,greater<pa> >q;
int main()
{
    n=read();
    for(int i=1;i<=n;++i)
    {
        int p=read();a[p]=i;
        if(p<i) return 0*puts("No");
        for(int j=1;j<i;++j) q.push(make_pair(p-j,i));
    }
    for(int i=1;i<=n*n;++i)
    {
        if(q.size()&&q.top().first<i) return 0*puts("No");
        if(a[i])
        {
            for(int j=a[i]+1;j<=n;++j) q.push(make_pair(n*n,a[i]));
            continue;
        }
        if(!q.size()) return 0*puts("No");
        a[i]=q.top().second,q.pop();
    }
    puts("Yes");
    for(int i=1;i<=n*n;++i) printf("%d ",a[i]);
    return 0;
}
