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, m, k, q, x;
  long long tab[10001];
  vector<array<long long, 3>> ans;
  long long answered[101];
  vector<array<long long, 3>> cands[101];
  void solve() {
    cin >> n >> k;
    x = (n - 1) / (k - 1) + 1;
    for (long long i = 1; i <= n * k; ++i) {
      cin >> tab[i];
    }
    for (long long col = 1; col <= n; ++col) answered[col] = 0;
    for (long long col = 1; col <= n; ++col) {
      long long lastInd = -1;
      long long which = 0;
      for (long long i = 1; i <= n * k; ++i) {
        if (tab[i] == col) {
          which++;
          if (which != 1) {
            cands[which - 2].push_back({i, col, lastInd});
          }
          lastInd = i;
        }
      }
    }
    for (long long i = 0; i < k - 1; ++i) {
      sort(cands[i].begin(), cands[i].end());
      long long stillLeft = x;
      for (long long j = 0; j < cands[i].size(); ++j) {
        if (!stillLeft) break;
        if (answered[cands[i][j][1]] == 0) {
          stillLeft--;
          answered[cands[i][j][1]] = 1;
          ans.push_back({cands[i][j][1], cands[i][j][2], cands[i][j][0]});
        }
      }
    }
    sort(ans.begin(), ans.end());
    for (auto &i : ans) {
      cout << i[1] << " " << i[2] << endl;
    }
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    long long z = 1;
    while (z--) {
      solve();
    }
  }
  