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: #include <bits/stdc++.h>
  struct interval {
    int l, r;
    int operator<(const interval &i) const { return r < i.r; }
  } is[10005];
  int n, k, a[10005], tot, app[105][105], cnt[10005], ci, lim, sel[105];
  int sell[105], selr[105];
  int main() {
    scanf("%d%d", &n, &k);
    tot = n * k;
    lim = (n + k - 2) / (k - 1);
    for (int i = 1; i <= tot; ++i) {
      scanf("%d", a + i);
      app[a[i]][++cnt[a[i]]] = i;
    }
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j < k; ++j) is[++ci] = (interval){app[i][j], app[i][j + 1]};
    std::sort(is + 1, is + ci + 1);
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= ci; ++i)
      if (!sel[a[is[i].l]]) {
        int can = 1;
        for (int j = is[i].l; j <= is[i].r; ++j) can &= cnt[j] < lim;
        if (can) {
          for (int j = is[i].l; j <= is[i].r; ++j) ++cnt[j];
          sell[a[is[i].l]] = is[i].l;
          selr[a[is[i].r]] = is[i].r;
          sel[a[is[i].l]] = 1;
        }
      }
    for (int i = 1; i <= n; ++i) printf("%d %d\n", sell[i], selr[i]);
    return 0;
  }
  