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;
  vector<array<int, 2>> solve_case(int balls, vector<int> &arr) {
    vector<bool> alive(balls - 1, true);
    vector<array<int, 2>> answer(balls - 1);
    int cur = 0;
    for (int i = 0; i < balls - 1; i++) {
      map<int, vector<int>> cnt;
      int found = -1;
      while (true) {
        if (alive[arr[cur]]) {
          cnt[arr[cur]].push_back(cur);
          if (cnt[arr[cur]].size() > 1) {
            found = arr[cur];
            break;
          }
        }
        cur += 1;
      }
      assert(found + 1);
      answer[found][0] = cnt[found][0];
      answer[found][1] = cnt[found][1];
      alive[found] = false;
      cur += 1;
    }
    return answer;
  }
  void solve_single(int balls, vector<array<int, 2>> &arr, int offset,
                    vector<array<int, 2>> &answer) {
    vector<int> single(balls * (balls - 1));
    for (int i = 0; i < balls * (balls - 1); i++) {
      single[i] = arr[i][0];
    }
    vector<array<int, 2>> ans = solve_case(balls, single);
    for (int i = 0; i < balls - 1; i++) {
      answer[offset + i][0] = arr[ans[i][0]][1];
      answer[offset + i][1] = arr[ans[i][1]][1];
    }
  }
  int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    int n_orig = n;
    vector<int> arr(n * k);
    for (int i = 0; i < n * k; i++) {
      scanf("%d", &arr[i]);
      arr[i] -= 1;
    }
    while (n % (k - 1) != 0) {
      for (int i = 0; i < k; i++) {
        arr.push_back(n);
      }
      n += 1;
    }
    int step = k - 1;
    vector<array<int, 2>> answer(n);
    for (int col = 0; col < n; col += step) {
      int col_start = col, col_end = col + step;
      vector<array<int, 2>> solve;
      for (int i = 0; i < n * k; i++) {
        if (col_start <= arr[i] && arr[i] < col_end) {
          solve.push_back({arr[i] - col_start, i});
        }
      }
      assert((int)solve.size() == k * (k - 1));
      solve_single(k, solve, col_start, answer);
    }
    for (int i = 0; i < n_orig; i++) {
      printf("%d %d\n", answer[i][0] + 1, answer[i][1] + 1);
    }
    return 0;
  }
  