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 INF = 1 << 30;
  const long long int MAX = 1e9 + 7;
  void array_show(int* array, int array_n, char middle = ' ') {
    for (int i = 0; i < array_n; i++)
      printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
  }
  void array_show(long long int* array, int array_n, char middle = ' ') {
    for (int i = 0; i < array_n; i++)
      printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
  }
  void array_show(vector<int>& vec_s, int vec_n = -1, char middle = ' ') {
    if (vec_n == -1) vec_n = vec_s.size();
    for (int i = 0; i < vec_n; i++)
      printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
  }
  void array_show(vector<long long int>& vec_s, int vec_n = -1,
                  char middle = ' ') {
    if (vec_n == -1) vec_n = vec_s.size();
    for (int i = 0; i < vec_n; i++)
      printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
  }
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v1) {
    int n = v1.size();
    for (int i = 0; i < n; i++) {
      if (i) os << " ";
      os << v1[i];
    }
    return os;
  }
  template <typename T1, typename T2>
  ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
    os << p.first << " " << p.second;
    return os;
  }
  template <typename T>
  istream& operator>>(istream& is, vector<T>& v1) {
    int n = v1.size();
    for (int i = 0; i < n; i++) is >> v1[i];
    return is;
  }
  template <typename T1, typename T2>
  istream& operator>>(istream& is, pair<T1, T2>& p) {
    is >> p.first >> p.second;
    return is;
  }
  namespace sol {
  void solve() {
    int n, m;
    int i, j, k;
    int a, b, c;
    cin >> n >> m;
    vector<int> v1, cnt(n), bef(n);
    vector<char> used(n);
    vector<pair<int, int> > vs(n);
    for (i = 0; i < n * m; i++) {
      cin >> a;
      v1.push_back(a - 1);
    }
    for (i = 0; i < (n - 1) / (m - 1) + 1; i++) {
      int l = i * (m - 1), r = min((i + 1) * (m - 1), n);
      a = 2;
      for (j = 0; j < n * m; j++) {
        if (v1[j] < l || r <= v1[j]) continue;
        cnt[v1[j]]++;
        if (used[v1[j]]) continue;
        if (cnt[v1[j]] != a) {
          bef[v1[j]] = j;
          continue;
        }
        used[v1[j]] = 1;
        vs[v1[j]] = {bef[v1[j]] + 1, j + 1};
        a++;
      }
    }
    for (i = 0; i < n; i++) {
      cout << vs[i] << endl;
    }
  }
  }  // namespace sol
  int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    sol::solve();
  }
  