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 <algorithm>
using namespace std;
typedef pair<int,int> P;
#define F first
#define S second
int n;
int p[250005];
P a[505];
int main(void){
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i].F);
        a[i].S=i+1;
    }
    sort(a,a+n);
    int q=1;
    for(int i=0;i<n;i++){
        int v=a[i].S,o=a[i].F;
        p[o]=v;
        for(int j=0;j<v-1;j++){
            while(p[q]>0)q++;
            if(q>o){
                printf("No\n");
                return 0;
            }
            p[q]=v;
        }
    }
    q=1;
    for(int i=0;i<n;i++){
        int v=a[i].S,o=a[i].F,s=n-v;
        for(int j=0;j<s;j++){
            while(p[q]>0)q++;
            if(q<o){
                printf("No\n");
                return 0;
            }
            p[q]=v;
        }
    }
    printf("Yes\n");
    for(int i=1;i<=n*n;i++){
        printf("%d",p[i]);
        if(i!=n*n)printf(" ");
    }
    printf("\n");
}