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;
  namespace debug {
  template <class c>
  struct rge {
    c b, e;
  };
  template <class c>
  rge<c> range(c i, c j) {
    return rge<c>{i, j};
  }
  template <class c>
  char elo(...);
  template <class c>
  auto elo(c* a) -> decltype(cerr << *a, 0);
  struct stream {
    ~stream() { cerr << endl; }
    template <class c>
    typename enable_if<sizeof elo<c>(0) != 1, stream&>::type operator<<(c i) {
      cerr << boolalpha << i;
      return *this;
    }
    template <class c>
    typename enable_if<sizeof elo<c>(0) == 1, stream&>::type operator<<(c i) {
      return *this << range(begin(i), end(i));
    }
    template <class a, class b>
    stream& operator<<(pair<a, b> p) {
      return *this << "(" << p.first << ", " << p.second << ")";
    }
    template <class c>
    stream& operator<<(rge<c> d) {
      *this << "[";
      for (auto it = d.b; it != d.e; it++) *this << ", " + 2 * (it == d.b) << *it;
      return *this << "]";
    }
    stream& _dbg(const string& s, int i, int b) { return *this; }
    template <class c, class... cs>
    stream& _dbg(const string& s, int i, int b, c arg, cs... args) {
      if (i == s.size()) return (*this << ": " << arg << "");
      b += (s[i] == '(') + (s[i] == '[') + (s[i] == '{');
      b -= (s[i] == ')') + (s[i] == ']') + (s[i] == '}');
      if (s[i] == ',' && b == 0)
        return (*this << ": " << arg << "     ")._dbg(s, i + 1, b, args...);
      return (s[i] == ' ' ? *this : *this << s[i])
          ._dbg(s, i + 1, b, arg, args...);
    }
  };
  }  // namespace debug
  template <class p, class q>
  pair<p, q> operator-(pair<p, q> a, pair<p, q> b) {
    return make_pair(a.first - b.first, a.second - b.second);
  }
  template <class p, class q>
  pair<p, q> operator+(pair<p, q> a, pair<p, q> b) {
    return make_pair(a.first + b.first, a.second + b.second);
  }
  template <class p, class q>
  void umin(p& a, const q& b) {
    if (a > b) a = b;
  }
  template <class p, class q>
  void umax(p& a, const q& b) {
    if (a < b) a = b;
  }
  using lf = long double;
  using ll = long long;
  using cll = const ll;
  using cint = const int;
  using pll = pair<ll, ll>;
  using pii = pair<int, int>;
  const int N = 1000005;
  int t[N];
  int last[N];
  bool done[N];
  int cnt[N];
  pair<int, int> out[N];
  void solve() {
    int n, i, j, k;
    scanf("%d%d", &n, &k);
    vector<pair<int, int> > vec;
    for (i = 1; i <= n * k; i++) {
      scanf("%d", &t[i]);
      if (last[t[i]] != 0) vec.push_back({i, last[t[i]]});
      last[t[i]] = i;
    }
    int lim = (n + k - 2) / (k - 1);
    sort(vec.begin(), vec.end());
    for (pair<int, int> p : vec) {
      int a = p.second;
      int b = p.first;
      int col = t[a];
      assert(t[a] == t[b]);
      if (done[col] == true) continue;
      for (i = a; i <= b; i++)
        if (cnt[i] == lim) break;
      if (i <= b) continue;
      for (i = a; i <= b; i++) cnt[i] += 1;
      out[col] = {a, b};
      done[col] = true;
    }
    for (i = 1; i <= n; i++) printf("%d %d\n", out[i].first, out[i].second);
  }
  int main() {
    int q = 1;
    while (q--) solve();
    return 0;
  }
  