Problem: 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

Solution (in PYTHON3):

Solution:
# Problem code: 509B

# n = number of piles, k = number of colors
n, k = input().split()
n, k = int(n), int(k)

# there are n amounts. These are number of pebbles in each pile
amounts = [int(i) for i in input().split()]

# if the difference between the biggest stack and the smallest stack is bigger than the number of
# colors, then there are not enough colors to fill the difference and make those stacks fulfill the
# requirements
if (max(amounts) - min(amounts)) > k:
    print("NO")
else:
    print("YES")

    # for each pile, divide the amount up into how many of each color, and fill them up with that
    # many of each color. Then, with the leftover amount, add the first few colors of the color
    # list (1,2,3,4...) to the end. Basically this is just painting the pebbles alternating colors
    # for example if there are 3 colors one pile might be 1 2 3 1 2 3 1 2 3 1 2 but this algorithm
    # for speed sake will combine colors and make 1 1 1 1 2 2 2 2 3 3 3
    for amount in amounts:
        count = amount // k
        leftover = amount % k
        for i in range(k):
            print((str(i+1)+" ")*count, end='')
        print(' '.join([str(i+1) for i in range(leftover)]))