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;
  void enable_comma() {}
  string tostring(char c) {
    string s = "";
    s += c;
    return s;
  }
  string tostring(string s) { return "\"" + s + "\""; }
  string tostring(const char *c) { return tostring((string)c); }
  string tostring(long long x) {
    if (x < 0) return "-" + tostring(-x);
    if (x > 9)
      return tostring(x / 10) + tostring(char('0' + x % 10));
    else
      return tostring(char('0' + x));
  }
  string tostring(int x) { return tostring((long long)x); }
  string tostring(unsigned long long x) {
    if (x > 9)
      return tostring((long long)(x / 10)) + tostring(char('0' + x % 10));
    else
      return tostring(char('0' + x));
  }
  string tostring(unsigned x) { return tostring((long long)x); }
  string tostring(double x) {
    static char res[114];
    sprintf(res, "%lf", x);
    string s = tostring(res);
    return s.substr(1, (int)s.size() - 2);
  }
  string tostring(long double x) { return tostring((double)x); }
  template <class A, class B>
  string tostring(pair<A, B> p) {
    return "(" + tostring(p.first) + "," + tostring(p.second) + ")";
  }
  template <class T>
  string tostring(T v) {
    string res = "";
    for (auto p : v) res += (res.size() ? "," : "{") + tostring(p);
    return res.size() ? res + "}" : "{}";
  }
  template <class A>
  string tostring(A *a, int L, int R) {
    return tostring(vector<A>(a + L, a + R + 1));
  };
  template <class A>
  string tostring(A a, int L, int R) {
    return tostring(a.data(), L, R);
  }
  string tostrings() { return ""; }
  template <typename Head, typename... Tail>
  string tostrings(Head H, Tail... T) {
    return tostring(H) + " " + tostrings(T...);
  }
  long long read() {
    long long x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f = ch == '-', ch = getchar();
    while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    return f ? -x : x;
  }
  template <class T>
  void ckmax(T &x, const T y) {
    if (x < y) x = y;
  }
  template <class T>
  void ckmin(T &x, const T y) {
    if (x > y) x = y;
  }
  const int N = 105;
  int n, k;
  int a[N * N], cnt[N * N];
  int vis[N], c[N];
  pair<int, int> ans[N];
  int main() {
    n = read(), k = read();
    int lim = (n + k - 2) / (k - 1);
    for (int i = (1); i <= (n * k); i++) a[i] = read();
    int p = 1;
    for (int i = (1); i <= (n); i++) {
      memset(c, 0, sizeof(c));
      int flag = 0;
      for (int j = (p); j <= (n * k); j++) {
        if (vis[a[j]]) continue;
        if (c[a[j]]) {
          for (int k = (c[a[j]]); k <= (j); k++) {
            cnt[k]++;
            if (cnt[k] == lim) ckmax(p, k + 1);
          }
          ans[a[j]] = make_pair(c[a[j]], j);
          vis[a[j]] = 1;
          flag = 1;
          break;
        }
        c[a[j]] = j;
      }
      (!(flag) ? cout << "Assertion failed!" << endl
                      << "function: " << __FUNCTION__ << endl
                      << "line: " << 167 << endl
                      << "expression: "
                      << "flag" << endl,
       exit(3), 0 : 1);
    }
    for (int i = (1); i <= (n); i++)
      printf("%d %d\n", ans[i].first, ans[i].second);
    return 0;
  }
  