There are n piles of pebbles on the table, the i-th pile contains ai pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that bi, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |bi, c - bj, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then bi, c is considered to be zero.

Input

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100) denoting number of pebbles in each of the piles.

Output

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain ai space-separated integers. j-th (1 ≤ j ≤ ai) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.

Examples

Input

4 4
1 2 3 4


Output

YES
1
1 4
1 2 4
1 2 3 4


Input

5 2
3 2 4 1 3


Output

NO


Input

5 4
3 2 4 3 5


Output

YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4
def yes_or_no(colors, color_count):
    last = colors[-1]
    first = colors[0]
    for i in range(color_count):
        if not (-2 < last[0][i] - first[0][i] < 2):
            return "NO"
    return "YES"


if __name__ == '__main__':
    color_count = int(input().split()[1])
    piles = list(map(int, input().split()))
    for i in range(len(piles)):
        piles[i] = (piles[i], i)
    new_piles = sorted(piles)
    colors = []
    for i in range(len(new_piles)):
        pile = new_piles[i][0]
        default = pile // color_count + 1
        more_counts = pile % color_count
        pile_colors = [default] * more_counts
        pile_colors += [default - 1] * (color_count - more_counts)
        colors.append((pile_colors, new_piles[i][1]))
    yes_no = yes_or_no(colors, color_count)
    print(yes_no)
    if yes_no == "YES":
        for vector in [item[0] for item in sorted(colors, key=lambda x: x[1])]:
            for i in range(len(vector)):
                count = vector[i]
                for j in range(count):
                    print(i + 1, end=" ")
            print()





