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 <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 505*505;
struct Node{
    int pos,num;
    bool operator < (const Node &b)const
    {
        return pos<b.pos;
    }
}a[505];
int t[MAXN],n;
inline void fail()
{
    puts("No");
    exit(0);
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;++i)scanf("%d",&a[i].pos),a[i].num=i;
    sort(a+1,a+n+1);
    int p=1;
    for(int i=1;i<=n;++i){
        if(a[i].pos==a[i-1].pos)fail();
        int tmp=a[i].num-1;
        while(tmp--){while(t[p])++p;if(p>=a[i].pos)fail();t[p++]=a[i].num;}
        t[a[i].pos]=a[i].num;
    }
    p=n*n;
    for(int i=n;i>0;--i){
        int tmp=n-a[i].num;
        while(tmp--){while(t[p])--p;if(p<=a[i].pos)fail();t[p--]=a[i].num;}
    }
    puts("Yes");
    for(int i=1;i<=n*n;++i)printf("%d ",t[i]);
    return 0;
}
