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 double pi = 3.14159265358979323;
  long long int n, k, m, tc, ans = 0, count = 0, mymax = -1e18, mymin = 1e18, x,
                             y;
  string final_ans = "";
  string s = "";
  template <typename T>
  ostream& operator<<(ostream& os, const vector<T>& v) {
    for (int i = 0; i < v.size(); i += 1) os << v[i] << " ";
    os << '\n';
    return os;
  }
  long long int fastexp(long long int base, long long int exp,
                        long long int mod) {
    long long int res = 1;
    while (exp > 0) {
      if (exp % 2) {
        res *= base;
        res %= mod;
      }
      exp /= 2;
      base *= base;
      base %= mod;
    }
    return res % mod;
  }
  long long int get_ans(vector<pair<int, int> > chords) {
    long long int ans = 0;
    for (int i = 0; i < chords.size(); i++) {
      for (int j = i + 1; j < chords.size(); j++) {
        long long int c11 = chords[i].first, c12 = chords[i].second;
        long long int c21 = chords[j].first, c22 = chords[j].second;
        if (c11 > c12) swap(c11, c12);
        if (c21 > c22) swap(c21, c22);
        vector<long long int> new_vec = {c11, c12, c21, c22};
        vector<long long int> one_vec = {c11, c21, c12, c22};
        vector<long long int> two_vec = {c21, c11, c22, c12};
        sort(new_vec.begin(), new_vec.end());
        if (new_vec == one_vec || new_vec == two_vec) ans++;
      }
    }
    return ans;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> n >> k;
    vector<long long int> vec(n * k);
    vector<long long int> cnts(n * k, 0);
    vector<pair<long long int, long long int> > matched(n + 1, {-1, -1});
    for (int j = 0; j < n * k; j += 1) cin >> vec[j];
    vector<int> currStack;
    long long int max_cnt = (n + k - 2) / (k - 1);
    for (int j = 0; j < n * k; j += 1) {
      long long int toFind = vec[j];
      if (matched[toFind].first != -1) continue;
      long long int found_match = -1;
      for (int itr = j - 1; itr >= 0; itr--) {
        if (cnts[itr] == max_cnt)
          break;
        else if (vec[itr] == toFind) {
          for (int mystart = itr; mystart <= j; mystart++) cnts[mystart]++;
          found_match = itr;
        }
      }
      if (found_match != -1) matched[toFind] = {found_match + 1, j + 1};
    }
    for (int j = 0; j < n; j += 1) {
      cout << matched[j + 1].first << " " << matched[j + 1].second << '\n';
    }
  }
  