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
  input = sys.stdin.readline
  from heapq import *
  
  def main():
      n, k = map(int, input().split())
      clst = list(map(int, input().split()))
      max_ = (n + k - 1 - 1) // (k - 1)
      cnt = [0] * (n * k)
      aft = [-1] * n
      nex = [-1] * (n * k)
      bef = [-1] * (n * k)
      for i in range(n * k - 1, -1, -1):
          c = clst[i] - 1
          nex[i] = aft[c]
          aft[c] = i
      used = [False] * n
      hq = []
      for i in range(n * k):
          c = clst[i] - 1
          if not used[c]:
              used[c] = True
              j = nex[i]
              bef[j] = i
              heappush(hq, -j)
              for l in range(i, j + 1):
                  cnt[l] += 1
          
          while cnt[i] > max_:
              j = -heappop(hq)
              used[clst[j] - 1] = False
              i = bef[j]
              for l in range(i, j + 1):
                  cnt[l] -= 1
      ans = [-1] * n
      for h in hq:
          h = -h
          ans[clst[h] - 1] = (bef[h] + 1, h + 1)
      for row in ans:
          print(*row)
              
      
      
          
  for _ in range(1):
      main()