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;
  template <typename T1, typename T2>
  inline void mine(T1 &x, const T2 &y) {
    if (y < x) x = y;
  }
  template <typename T1, typename T2>
  inline void maxe(T1 &x, const T2 &y) {
    if (x < y) x = y;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &b) {
    for (auto &k : b) os << k << ' ';
    return os;
  }
  template <typename T>
  istream &operator>>(istream &is, vector<T> &b) {
    for (auto &k : b) is >> k;
    return is;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long N = 101;
  long long last[N];
  long long a[N * N];
  pair<long long, long long> ans[N];
  int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long n, k;
    cin >> n >> k;
    memset(last, -1, sizeof(last));
    long long can = (n + (k - 2)) / (k - 1);
    for (long long i = 0; i < n; ++i) ans[i] = {-1, -1};
    for (long long i = 0; i < n * k; ++i) {
      long long x;
      cin >> x;
      --x;
      if (last[x] == -1) {
        last[x] = i;
      } else if (ans[x].first == -1) {
        long long mx = 0;
        for (long long j = last[x]; j <= i; ++j) {
          maxe(mx, a[j]);
        }
        if (mx < can) {
          for (long long j = last[x]; j <= i; ++j) {
            ++a[j];
          }
          ans[x] = {last[x] + 1, i + 1};
        }
        last[x] = i;
      }
    }
    for (long long i = 0; i < n; ++i) {
      cout << ans[i].first << ' ' << ans[i].second << '\n';
    }
  }
  