description: The numbers 1,   2,   ...,   n ⋅ k are colored with n colors. These colors are indexed by 1,   2,   ...,   n. For each 1 ≤ i ≤ n, there are exactly k numbers colored with color i.
  
  Let [a,   b] denote the interval of integers between a and b inclusive, that is, the set \\{a,   a + 1,   ...,   b\}. You must choose n intervals [a_1,   b_1],   [a_2,   b_2],   ..., [a_n,   b_n] such that: 
  
    * for each 1 ≤ i ≤ n, it holds 1 ≤ a_i < b_i ≤ n ⋅ k; 
    * for each 1 ≤ i ≤ n, the numbers a_i and b_i are colored with color i; 
    * each number 1 ≤ x ≤ n ⋅ k belongs to at most \left⌈ (n)/(k - 1) \right⌉ intervals. 
  
  
  
  One can show that such a family of intervals always exists under the given constraints.
  
  Input
  
  The first line contains two integers n and k (1 ≤ n ≤ 100, 2 ≤ k ≤ 100) — the number of colors and the number of occurrences of each color.
  
  The second line contains n ⋅ k integers c_1,   c_2,   ...,   c_{nk} (1 ≤ c_j ≤ n), where c_j is the color of number j. It is guaranteed that, for each 1 ≤ i ≤ n, it holds c_j = i for exactly k distinct indices j.
  
  Output
  
  Output n lines. The i-th line should contain the two integers a_i and b_i.
  
  If there are multiple valid choices of the intervals, output any.
  
  Examples
  
  Input
  
  
  4 3
  2 4 3 1 1 4 2 3 2 1 3 4
  
  
  Output
  
  
  4 5
  1 7
  8 11
  6 12
  
  Input
  
  
  1 2
  1 1
  
  
  Output
  
  
  1 2
  
  
  Input
  
  
  3 3
  3 1 2 3 2 1 2 1 3
  
  
  Output
  
  
  6 8
  3 7
  1 4
  
  Input
  
  
  2 3
  2 1 1 1 2 2
  
  
  Output
  
  
  2 3
  5 6
  
  Note
  
  In the first sample, each number can be contained in at most \left⌈ (4)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:
  
  <image>
  
  In the second sample, the only interval to be chosen is forced to be [1,   2], and each number is indeed contained in at most \left⌈ (1)/(2 - 1) \right⌉ = 1 interval.
  
  In the third sample, each number can be contained in at most \left⌈ (3)/(3 - 1) \right⌉ = 2 intervals. The output is described by the following picture:
  
  <image>
solution: import sys
  import math
  
  if sys.subversion[0] == "PyPy":
      import io, atexit
  
      sys.stdout = io.BytesIO()
      atexit.register(lambda: sys.__stdout__.write(sys.stdout.getvalue()))
  
      sys.stdin = io.BytesIO(sys.stdin.read())
      input = lambda: sys.stdin.readline().rstrip()
  
  #t = int(raw_input())
  
  # python codesforces210710.py < input.txt
  #for x in range(t):
  str1 = raw_input()
  str2 = raw_input()
  n, k = str1.split(' ')
  li = str2.split(' ')
  
  #input
  n = int(n)
  k = int(k)
  li = map(int, li)
  #variables
  matt = []
  setdone = set()
  rr = list(range(1,n+1))
  biD = dict.fromkeys(rr)
  bigfDD = biD.copy()
  for i, l in enumerate(li):
      matt.append([i+1, l])
  
  #solve
  while len(matt) > 0:
      bigD = biD.copy()
      for i in range(0, len(matt)):
          if matt[i][1] not in setdone:
              if bigD.get(matt[i][1]):
                  bigfDD[matt[i][1]] = str(bigD.get(matt[i][1])) + ' ' + str(matt[i][0])
                  setdone.add(matt[i][1])
                  bigD = biD.copy()
              else:
                  bigD[matt[i][1]] = matt[i][0]
      matt = [x for x in matt if x[1] not in setdone]
  
  
  
  for i in range(1,n+1):
      print(bigfDD[i])