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;
  template <typename T>
  using MinHeap = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  using MaxHeap = priority_queue<T>;
  const long double PI = 3.14159265358979323846;
  const long double E = 2.718281828459;
  long long MOD = (1ll << 31) - 1;
  long long M(long long n) { return (n % MOD + MOD) % MOD; }
  long long pow2(long long n) { return 1ll << n; }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  int main(int argn, char **argv) {
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    cin.tie(nullptr);
    int T = 1;
    for (int(I) = (0); (I) <= ((T)-1); (I)++) {
      long long n, k;
      cin >> n >> k;
      vector<long long> pos[n + 1];
      for (int(p) = (0); (p) <= ((n * k) - 1); (p)++) {
        long long col;
        cin >> col;
        pos[col].emplace_back(p + 1);
      }
      vector<tuple<long long, long long, long long>> seg;
      for (int(col) = (1); (col) <= (n); (col)++) {
        for (int(i) = (0); (i) <= ((k - 1) - 1); (i)++) {
          seg.emplace_back(pos[col][i + 1], pos[col][i], col);
        }
      }
      sort((seg).begin(), (seg).end());
      long long tot = 0;
      vector<long long> done(n + 1, 0);
      vector<pair<long long, long long>> ans(n + 1);
      while (tot < n) {
        long long idx = 0, crt = 0;
        while (idx < seg.size()) {
          auto [r, l, col] = seg[idx++];
          if (l <= crt || done[col]) continue;
          done[col] = 1, tot++;
          ans[col] = {l, r};
          crt = r;
        }
      }
      for (int(col) = (1); (col) <= (n); (col)++) {
        auto [l, r] = ans[col];
        ;
        cout << l << " " << r << endl;
      }
    }
    return 0;
  }
  