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;
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n * k);
    for (long long& i : v) {
      cin >> i;
      --i;
    }
    vector<vector<long long>> pos(n);
    for (long long i = 0; i < n * k; ++i) pos[v[i]].push_back(i);
    vector<pair<long long, long long>> ans(n), rng;
    for (long long i = 0; i < n; ++i) {
      for (long long j = 0; j < k - 1; ++j) {
        rng.push_back({pos[i][j], pos[i][j + 1]});
      }
    }
    sort(rng.begin(), rng.end(),
         [&](pair<long long, long long> a, pair<long long, long long> b) {
           return a.second < b.second;
         });
    pair<long long, long long> zero = {0, 0};
    long long bound = (n + k - 2) / (k - 1);
    vector<long long> full(n * k);
    for (long long i = 0; i < rng.size(); ++i) {
      long long col = v[rng[i].first];
      if (ans[col] != zero) continue;
      long long y = 1;
      for (long long x = rng[i].first; x <= rng[i].second; ++x)
        y &= full[x] < bound;
      if (!y) continue;
      for (long long x = rng[i].first; x <= rng[i].second; ++x) ++full[x];
      ans[col] = rng[i];
    }
    assert(ans.back() != zero);
    for (auto [a, b] : ans) cout << a + 1 << " " << b + 1 << endl;
  }
  