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;
  const int mod = 998244353;
  const int maxn = 1005;
  void go() {
    int n, k;
    cin >> n >> k;
    vector<int> v[n + 1];
    for (int i = 1; i < n * k + 1; i++) {
      int a;
      cin >> a;
      v[a].push_back(i);
    }
    int fin[101] = {}, l[101], r[101];
    int iter = 0;
    while (1) {
      priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
      for (int i = 1; i < n + 1; i++) {
        if (!fin[i]) {
          vector<int> a{v[i][1], v[i][0], i, 1};
          pq.push(a);
        }
      }
      if (!pq.size()) break;
      int ok = 0;
      while (pq.size()) {
        auto now = pq.top();
        pq.pop();
        if (now[1] > ok) {
          fin[now[2]] = 1;
          l[now[2]] = now[1];
          r[now[2]] = now[0];
          ok = r[now[2]];
        } else {
          if (now[3] < k - 1) {
            now[3]++;
            now[0] = v[now[2]][now[3]];
            now[1] = v[now[2]][now[3] - 1];
            pq.push(now);
          }
        }
      }
    }
    for (int i = 1; i < n + 1; i++) {
      cout << l[i] << ' ' << r[i] << '\n';
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int c = 0;
    int t;
    if (!c) {
      t = 1;
    } else {
      cin >> t;
    }
    while (t--) {
      go();
    }
  }
  