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 max_n = 103, inf = 1000111222;
  int m[max_n * max_n];
  int cnt[max_n * max_n];
  int n, k;
  int mx;
  int lst[max_n];
  bool used[max_n];
  pair<int, int> res[max_n];
  int get_max(int from, int to) {
    int ans = cnt[from];
    for (int i = from + 1; i <= to; ++i) {
      ans = cnt[i] > ans ? cnt[i] : ans;
    }
    return ans;
  }
  void add1(int from, int to) {
    for (int i = from; i <= to; ++i) {
      ++cnt[i];
    }
  }
  vector<pair<int, int>> solve() {
    vector<pair<int, int>> ans;
    memset(used, 0, sizeof(used));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i <= n; ++i) {
      lst[i] = -1;
      res[i] = make_pair(-1, -1);
    }
    for (int i = 0; i < n * k; ++i) {
      int cur = m[i];
      if (used[cur]) {
        continue;
      }
      int pr = lst[cur];
      lst[cur] = i;
      if (pr == -1) {
        continue;
      }
      int tp = get_max(pr, i);
      if (tp == mx) {
        continue;
      }
      used[cur] = true;
      res[cur] = make_pair(pr, i);
      add1(pr, i);
    }
    for (int i = 0; i < n; ++i) {
      if (!used[i]) {
        return ans;
      }
    }
    for (int i = 0; i < n; ++i) {
      ans.push_back(res[i]);
    }
    return ans;
  }
  int main() {
    cin >> n >> k;
    for (int i = 0; i < n * k; ++i) {
      cin >> m[i];
      --m[i];
    }
    mx = (n + k - 2) / (k - 1);
    vector<pair<int, int>> ans = solve();
    if (ans.empty()) {
      reverse(m, m + n * k);
      ans = solve();
      if (ans.empty()) {
        exit(228);
      }
      for (int i = 0; i < n; ++i) {
        ans[i].first = (n * k - 1 - ans[i].first);
        ans[i].second = (n * k - 1 - ans[i].second);
        swap(ans[i].first, ans[i].second);
      }
    }
    for (int i = 0; i < n; ++i) {
      cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << "\n";
    }
    return 0;
  }
  