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 operator import itemgetter
  
  n,k=map(int,input().split())
  C=list(map(int,input().split()))
  
  CLIST=[[] for i in range(n)]
  
  for i in range(n*k):
      CLIST[C[i]-1].append(i)
  
  SUM=[0]*(n*k)
  
  
  MAX=(n+(k-2))//(k-1)
  
  C2=[]
  for i in range(n):
      for j in range(len(CLIST[i])-1):
          x,y=CLIST[i][j],CLIST[i][j+1]
          C2.append((x,y,i))
  
  C2.sort(key=itemgetter(1))
  
  seg_el=1<<((n*k+5).bit_length())
  SEG=[0]*(2*seg_el)
  LAZY=[0]*(2*seg_el)
  
  def indexes(L,R):
      INDLIST=[]
  
      R-=1
      
      L>>=1
      R>>=1
  
      while L!=R:
          if L>R:
              INDLIST.append(L)
              L>>=1
          else:
              INDLIST.append(R)
              R>>=1
  
      while L!=0:
          INDLIST.append(L)
          L>>=1
  
      return INDLIST
          
  
  def adds(l,r,x):
          
      L=l+seg_el
      R=r+seg_el
  
      L//=(L & (-L))
      R//=(R & (-R))
  
      UPIND=indexes(L,R)
      
      for ind in UPIND[::-1]:
          if LAZY[ind]!=0:
              plus_lazy=LAZY[ind]
              
              SEG[ind<<1]+=plus_lazy
              SEG[1+(ind<<1)]+=plus_lazy
              
              LAZY[ind<<1]+=plus_lazy
              LAZY[1+(ind<<1)]+=plus_lazy
              
              LAZY[ind]=0
      
      while L!=R:
          if L > R:
              SEG[L]+=x
              LAZY[L]+=x
              L+=1
              L//=(L & (-L))
  
          else:
              R-=1
              SEG[R]+=x
              LAZY[R]+=x
              R//=(R & (-R))
  
      for ind in UPIND:
          SEG[ind]=max(SEG[ind<<1],SEG[1+(ind<<1)])
          
  def getvalues(l,r):
  
      L=l+seg_el
      R=r+seg_el
  
      L//=(L & (-L))
      R//=(R & (-R))
  
      UPIND=indexes(L,R)
      
      for ind in UPIND[::-1]:
          if LAZY[ind]!=0:
              plus_lazy=LAZY[ind]
              
              SEG[ind<<1]+=plus_lazy
              SEG[1+(ind<<1)]+=plus_lazy
              
              LAZY[ind<<1]+=plus_lazy
              LAZY[1+(ind<<1)]+=plus_lazy
              
              LAZY[ind]=0
              
      ANS=0
  
      while L!=R:
          if L > R:
              ANS=max(ANS , SEG[L])
              L+=1
              L//=(L & (-L))
  
          else:
              R-=1
              ANS=max(ANS , SEG[R])
              R//=(R & (-R))
  
      return ANS
  
  ANS=[[-1,-1] for i in range(n)]
  
  for x,y,ind in C2:
      if ANS[ind]!=[-1,-1]:
          continue
  
      if getvalues(x,y)>=MAX:
          continue
  
      ANS[ind]=[x,y]
      adds(x,y+1,1)
  
  for x,y in ANS:
      print(x+1,y+1)
      
  
      
  
  
  