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;
  long long n, k;
  long long mas = 0;
  vector<pair<long long, long long>> ans(1000000);
  void solve(vector<vector<long long>> poses) {
    vector<long long> all;
    for (long long i = 0; i < poses.size(); i++) {
      for (long long j = 0; j < k; j++) {
        all.push_back(poses[i][j]);
      }
    }
    sort(all.begin(), all.end());
    map<long long, long long> conv;
    for (long long i = 0; i < all.size(); i++) {
      conv[all[i]] = i;
    }
    long long len = all.size();
    for (long long i = 0; i < poses.size(); i++) {
      for (long long j = 0; j < k; j++) {
        poses[i][j] = conv[poses[i][j]];
      }
    }
    vector<long long> start(poses.size());
    vector<bool> used(poses.size());
    for (long long _ = 0; _ < poses.size(); _++) {
      long long val = 1000000;
      long long ind = -1;
      for (long long i = 0; i < poses.size(); i++) {
        if (used[i]) {
          continue;
        }
        if (poses[i][start[i] + 1] < val) {
          val = poses[i][start[i] + 1];
          ind = i;
        }
      }
      ans[mas + ind].first = all[poses[ind][start[ind]]] + 1;
      ans[mas + ind].second = all[poses[ind][start[ind] + 1]] + 1;
      used[ind] = true;
      for (long long i = 0; i < poses.size(); i++) {
        if (used[i]) {
          continue;
        }
        if (poses[i][start[i]] < poses[ind][start[ind] + 1]) {
          start[i]++;
        }
      }
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    vector<vector<long long>> poses(n);
    for (long long i = 0; i < n * k; i++) {
      long long c;
      cin >> c;
      poses[c - 1].push_back(i);
    }
    for (long long i = 0; i < n; i += k - 1) {
      vector<vector<long long>> kek;
      for (long long j = i; j < min(i + k - 1, n); j++) {
        kek.push_back(poses[j]);
      }
      mas = i;
      solve(kek);
    }
    for (long long i = 0; i < n; i++) {
      cout << ans[i].first << " " << ans[i].second << '\n';
    }
  }
  