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;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long n, k;
    cin >> n >> k;
    vector<long long> c(n * k + 1);
    for (long long z = 1; (1 > 0 and z < n * k + 1) or (1 < 0 and z > n * k + 1);
         z += 1)
      cin >> c[z];
    ;
    vector<pair<long long, long long>> ivl(n + 1, {-1, -1});
    vector<pair<vector<long long>, long long>> a(n, {vector<long long>(), 0});
    for (long long i = 1; (1 > 0 and i < n * k + 1) or (1 < 0 and i > n * k + 1);
         i += 1)
      a[c[i] - 1].first.push_back(i);
    for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
      a[i].second = i + 1;
    long long t = 1;
    function<bool(pair<vector<long long>, long long>,
                  pair<vector<long long>, long long>)>
        comp = [&](pair<vector<long long>, long long> a1,
                   pair<vector<long long>, long long> a2) {
          return a1.first[t] < a2.first[t];
        };
    for (; t < k; ++t) {
      long long m = a.size();
      sort(a.begin(), a.end(), comp);
      for (long long i = 0;
           (1 > 0 and i < min(m, n / (k - 1) + (n % (k - 1) > 0))) or
           (1 < 0 and i > min(m, n / (k - 1) + (n % (k - 1) > 0)));
           i += 1)
        ivl[a[i].second] = {a[i].first[t - 1], a[i].first[t]};
      if (m > n / (k - 1) + (n % (k - 1) > 0)) {
        vector<pair<vector<long long>, long long>> b;
        for (long long i = n / (k - 1) + (n % (k - 1) > 0);
             (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
          b.push_back(a[i]);
        a = b;
      } else
        break;
    }
    for (long long i = 1; (1 > 0 and i < n + 1) or (1 < 0 and i > n + 1); i += 1)
      cout << ivl[i].first << " " << ivl[i].second << endl;
    return 0;
  }
  