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>
  #pragma GCC optimize("-Ofast")
  #pragma GCC target( \
      "sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
  #pragma GCC optimize("unroll-loops")
  using namespace std;
  template <typename T>
  void chkmax(T &first, T second) {
    first < second ? first = second : T();
  }
  template <typename T>
  void chkmin(T &first, T second) {
    second < first ? first = second : T();
  }
  template <typename T>
  void readint(T &first) {
    first = 0;
    int f = 1;
    char c;
    for (c = getchar(); !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    for (; isdigit(c); c = getchar()) first = first * 10 + (c - '0');
    first *= f;
  }
  template <typename T>
  T divceil(T first, T second) {
    return (first - 1) / second + 1;
  }
  int n, k, c[10005], last[105], cnt[105], f[105];
  struct interval {
    interval(int _c, int _l, int _r) : c(_c), l(_l), r(_r) {}
    interval() = default;
    int c, l, r;
  };
  vector<interval> V[105];
  interval ans[105];
  bool cmp(interval &i1, interval &i2) { return i1.r < i2.r; }
  int main() {
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
      last[i] = -1;
      cnt[i] = 0, f[i] = 0;
    }
    for (int i = 0; i < n * k; ++i) {
      readint(c[i]), --c[i];
      if (last[c[i]] >= 0) {
        V[cnt[c[i]]++].push_back(interval(c[i], last[c[i]] + 1, i + 1));
      }
      last[c[i]] = i;
    }
    memset(f, 0, sizeof(f));
    int b = divceil(n, k - 1);
    for (int i = 0; i < k; ++i) {
      sort(V[i].begin(), V[i].end(), cmp);
      int n_selected = 0;
      for (auto &itv : V[i]) {
        if (!f[itv.c] && n_selected < b) {
          ans[itv.c] = itv;
          f[itv.c] = 1;
          ++n_selected;
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      printf("%d %d\n", ans[i].l, ans[i].r);
    }
    return 0;
  }
  