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;
  using ll = long long;
  using pii = pair<long long, long long>;
  const double EPS = 1e-9;
  const long long INF = 1e18;
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << "(" << p.first << ", " << p.second << ")";
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  template <typename T>
  void debug_out(string s, T t) {
    cout << "[" << s << ": " << t << "]\n";
  }
  template <typename T, typename... U>
  void debug_out(string s, T t, U... u) {
    long long w = 0, c = 0;
    while (w < (long long)s.size()) {
      if (s[w] == '(') c++;
      if (s[w] == ')') c--;
      if (!c and s[w] == ',') break;
      w++;
    }
    cout << "[" << s.substr(0, w) << ": " << t << "] ";
    debug_out(s.substr(w + 2, (long long)s.size() - w - 1), u...);
  }
  const long long N = 1e6 + 5;
  vector<long long> idx[105];
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long n, k;
    cin >> n >> k;
    vector<long long> c(n * k + 1), prev(n * k + 1), done(n + 1);
    vector<pii> ans(n + 1);
    for (long long i = 1; i <= n * k; i++) cin >> c[i], idx[c[i]].push_back(i);
    bool flg = 1;
    while (true) {
      bool found = 0;
      for (long long &x : prev) x = 0;
      for (long long i = 1; i <= n * k; i++) {
        if (!done[c[i]]) {
          if (prev[c[i]] > 0) {
            done[c[i]] = 1, ans[c[i]].first = prev[c[i]], ans[c[i]].second = i,
            found = 1;
            for (long long &x : prev) x = 0;
          } else
            prev[c[i]] = i;
        }
      }
      if (!found) break;
    }
    for (long long i = 1; i <= n; i++)
      cout << ans[i].first << " " << ans[i].second << "\n";
    return 0;
  }
  