One day Ms Swan bought an orange in a shop. The orange consisted of *n*·*k* segments, numbered with integers from 1 to *n*·*k*. 

There were *k* children waiting for Ms Swan at home. The children have recently learned about the orange and they decided to divide it between them. For that each child took a piece of paper and wrote the number of the segment that he would like to get: the *i*-th (1 ≤ *i* ≤ *k*) child wrote the number *a*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ *n*·*k*). All numbers *a*<sub class="lower-index">*i*</sub> accidentally turned out to be different.

Now the children wonder, how to divide the orange so as to meet these conditions:
 -  each child gets exactly *n* orange segments; -  the *i*-th child gets the segment with number *a*<sub class="lower-index">*i*</sub> for sure; -  no segment goes to two children simultaneously. 
Help the children, divide the orange and fulfill the requirements, described above.
The first line contains two integers *n*, *k* (1 ≤ *n*, *k* ≤ 30). The second line contains *k* space-separated integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*k*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ *n*·*k*), where *a*<sub class="lower-index">*i*</sub> is the number of the orange segment that the *i*-th child would like to get.

It is guaranteed that all numbers *a*<sub class="lower-index">*i*</sub> are distinct.
# Time complexity: O(N*K) => Linear time complexity
# Space complexity: O(K) => Constant complexity
def solve(n, k, a):
    b = sorted(a)
    a_idx = 0
    k_idx = 0
    for i in range(1, n*k+1):
        if i != b[a_idx][0]:
            if len(a[k_idx]) == n:
                k_idx += 1
            a[k_idx].append(i)
        elif a_idx < len(a) -1:
            a_idx += 1
    return a

def print_res(a):
    for e in a:
        print(' '.join([str(x) for x in e]))

n, k = [int(x) for x in input().split()]
a = [[int(x)] for x in input().split()]
print_res(solve(n, k, a))