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>
  using namespace std;
  const int N = 10111;
  int n, K, m;
  int a[N];
  vector<int> c[105], d[105];
  int ans[105][2];
  bool vis[111];
  int main() {
    scanf("%d%d", &n, &K);
    m = n * K;
    for (int i = 1; i <= m; i++) {
      scanf("%d", a + i);
      c[a[i]].push_back(i);
      d[a[i]].push_back(0);
    }
    int upper = (n + K - 2) / (K - 1);
    for (int i = 1; i <= n; i++) {
      int mina = 1e9, id = -1, p = 0;
      for (int j = 1; j <= n; j++)
        if (!vis[j]) {
          for (int k = 1; k < K; k++) {
            if (d[j][k] < upper) {
              if (mina > c[j][k]) {
                mina = c[j][k];
                id = j;
                p = k;
              }
              break;
            }
          }
        }
      assert(id != -1);
      vis[id] = 1;
      ans[id][0] = c[id][p - 1];
      ans[id][1] = c[id][p];
      for (int j = 1; j <= n; j++) {
        for (int k = 1; k < K; k++) {
          if (c[id][p - 1] <= c[j][k] && c[id][p] >= c[j][k - 1]) {
            d[j][k]++;
          }
        }
      }
    }
    for (int i = 1; i <= n; i++) {
      printf("%d %d\n", ans[i][0], ans[i][1]);
    }
  }
  