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>
using namespace std;
typedef pair<int, int> P;

int main()
{
    int n;
    cin >> n;
    P p[502];
    int b[250002]{0};
    for(int i = 0; i < n; i++){
        int x;
        cin >> x;
        x--;
        b[x] = i + 1;
        p[i] = P(x, i + 1);
    }
    sort(p, p + n);
    int k = 0;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < p[i].second - 1; j++){
            while(b[k]) k++;
            if(k > p[i].first){
                cout << "No" << endl;
                return 0;
            }
            b[k] = p[i].second;
            k++;
        }
    }
    k = n * n - 1;
    for(int i = n - 1; i >= 0; i--){
        for(int j = 0; j < n - p[i].second; j++){
            while(b[k]) k--;
            if(k < p[i].first){
                cout << "No" << endl;
                return 0;
            }
            b[k] = p[i].second;
            k--;
        }
    }
    cout << "Yes" << endl;
    for(int i = 0; i < n * n; i++) cout << b[i] << " ";
    cout << endl;
}