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<int> vec;
  int n, k;
  const int dydis = 101;
  int nxt[dydis * dydis + 10];
  vector<pair<int, int>> transform(vector<int> &mas, vector<int> &take) {
    vector<pair<int, int>> ret;
    int i = 0;
    for (auto &x : mas) {
      bool tk = 0;
      for (auto &y : take) {
        if (y == x) tk = 1;
      }
      if (tk) ret.push_back({x, i});
      i++;
    }
    return ret;
  }
  vector<pair<int, int>> transform(vector<pair<int, int>> &mas, int ign, int st) {
    vector<pair<int, int>> ret;
    pair<int, int> x;
    for (int i = st; i < mas.size(); i++) {
      x = mas[i];
      if (x.first != ign) {
        ret.push_back(x);
      }
    }
    return ret;
  }
  vector<pair<int, int>> fAns(int l, int r) {
    vector<int> st;
    for (int i = l; i <= r; i++) {
      st.push_back(i);
    }
    auto mas = transform(vec, st);
    vector<pair<int, int>> ret;
    for (int i = 0; i < st.size(); i++) {
      for (auto x : mas) {
      }
      pair<int, int> mn = {dydis * dydis, dydis * dydis};
      for (int j = 0; j < mas.size(); j++) {
        mn = min(mn, make_pair(nxt[mas[j].second], mas[j].second));
      }
      int tol = 0;
      for (int j = 0; j < mas.size(); j++) {
        if (mn.first == mas[j].second) tol = j;
      }
      ret.push_back({mn.second, mn.first});
      mas = transform(mas, vec[mn.first], tol + 1);
    }
    return ret;
  }
  int main() {
    srand(time(0));
    cin >> n >> k;
    int grupiu = (n / (k - 1)) + (n % (k - 1) != 0);
    int grupeje = (n) / grupiu;
    vector<int> mas(grupiu, grupeje);
    int sm = grupeje * grupiu;
    for (int i = 0; i < grupiu; i++) {
      if (sm != n) {
        mas[i]++;
        sm++;
      }
    }
    for (int i = 0; i < n * k; i++) {
      int a;
      cin >> a;
      vec.push_back(a - 1);
    }
    for (int i = 0; i < vec.size(); i++) {
      int nx = n * k;
      for (int j = i + 1; j < vec.size(); j++) {
        if (vec[j] == vec[i]) {
          nx = j;
          break;
        }
      }
      nxt[i] = nx;
    }
    int l = 0;
    vector<pair<int, pair<int, int>>> ans;
    for (int i = 0; i < mas.size(); i++) {
      auto st = fAns(l, l + mas[i] - 1);
      l += mas[i];
      for (auto x : st) ans.push_back({vec[x.first], x});
    }
    sort(ans.begin(), ans.end());
    for (auto x : ans) {
      cout << x.second.first + 1 << " " << x.second.second + 1 << "\n";
      if (vec[x.second.first] != vec[x.second.second]) {
        cout << "NO";
        return 0;
      }
    }
    return 0;
  }
  