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 io,os
  import heapq
  input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def main(t):
  
  
      n,k = map(int,input().split())
      arr = list(map(int,input().split()))
      indexes = [[] for i in range(n)]
      tot = n*k
      for i in range(tot):
          indexes[arr[i]-1].append(i)
  
  
  
      bound = (n-1)//(k-1) + 1
      ans = [[] for i in range(n)]
      visited = [0]*n
  
  
  
      for p in range(k-1):
          heap = []
          for c in range(n):
              heapq.heappush(heap,[ indexes[c][p+1],indexes[c][p], c ])
          rest = bound 
  
          while heap:
              ele = heapq.heappop(heap)
              [end,front,c] = ele
              if visited[c]: continue
              visited[c] = 1 
              ans[c] = [front,end]
              rest -= 1
              if rest==0: break
  
  
              
               
              
  
  
  
  
  
          
      
  #    print(ans)       
  
  
      for ele in ans:
          print(ele[0]+1,ele[1]+1)
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  
  T = 1 #int(input())
  t = 1
  while t<=T:
      main(t)
      t += 1
  