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:
n, k = map(int, input().split())
a = [int(x) for x in input().split()]
if max(a) - min(a) > k:
    print("NO")
else:
    items = []
    for i in range(len(a)):
        l = [x for x in range(1, a[i] + 1)]
        if max(l) > k:
            i = 0
            temp = 1
            while i < len(l):
                if temp > k:
                    temp = 1
                if l[i] > k:
                    l[i] = temp
                    temp += 1
                i += 1
            # print(l)
        items.append(l)
    flag = 0
    for i in range(len(items)):
        for j in range(i + 1, len(items)):
            freq1 = [0] * 100
            freq2 = [0] * 100
            for k in range(len(items[i])):
                freq1[items[i][k]] += 1
            for k in range(len(items[j])):
                freq2[items[j][k]] += 1
            ans = []
            for k in range(len(freq1)):
                ans.append(abs(freq1[k] - freq2[k]))
            """for i in range(len(freq1)):
                ans.append(abs(freq1[i] - freq2[i]))"""
            if max(ans) > 1:
                print("NO")
                flag = 1
                break
        if flag == 1:
            break
    if flag == 0:
        print("YES")
        for i in items:
            for j in i:
                print(j, end=" ")
            print()