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<bits/stdc++.h>
using namespace std;
const int N = 505;
int n, X[N], C[N], A[N * N];
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &X[i]);
        if (A[X[i]])
            return !printf("No\n");
        A[X[i]] = i;
    }
    for (int i = 1; i <= n * n; i++)
        if (!A[i])
        {
            int id = -1;
            for (int j = 1; j <= n; j++)
                if (C[j] < j - 1 && i < X[j] && (id == -1 || X[j] < X[id]))
                    id = j;
            if (id == -1)
                for (int j = 1; j <= n; j++)
                    if (C[j] >= j - 1 && C[j] < n - 1 && i > X[j])
                        id = j;
            if (id == -1)
                return !printf("No\n");
            A[i] = id;
            C[A[i]] ++;
        }
    printf("Yes\n");
    for (int i = 1; i <= n * n; i++)
        printf("%d ", A[i]);
    return 0;
}