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>
  void chkmin(T1 &x, T2 y) {
    if (x > y) x = y;
  }
  template <typename T1, typename T2>
  void chkmax(T1 &x, T2 y) {
    if (x < y) x = y;
  }
  namespace fastio {
  char rbuf[1 << 23], *p1 = rbuf, *p2 = rbuf, wbuf[1 << 23], *p3 = wbuf;
  inline char getc() {
    return p1 == p2 &&
                   (p2 = (p1 = rbuf) + fread(rbuf, 1, 1 << 23, stdin), p1 == p2)
               ? -1
               : *p1++;
  }
  inline void putc(char x) { (*p3++ = x); }
  template <typename T>
  void read(T &x) {
    x = 0;
    char c = getchar();
    T neg = 0;
    while (!isdigit(c)) neg |= !(c ^ '-'), c = getchar();
    while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
    if (neg) x = (~x) + 1;
  }
  template <typename T>
  void recursive_print(T x) {
    return (!x) ? void() : (recursive_print(x / 10), putc(x % 10 ^ 48), void());
  }
  template <typename T>
  void print(T x) {
    (!x) && (putc('0'), 0);
    (x < 0) && (putc('-'), x = ~x + 1);
    recursive_print(x);
  }
  template <typename T>
  void print(T x, char c) {
    (!x) && (putc('0'), 0);
    (x < 0) && (putc('-'), x = ~x + 1);
    recursive_print(x);
    putc(c);
  }
  void print_final() { fwrite(wbuf, 1, p3 - wbuf, stdout); }
  }  // namespace fastio
  const int MAXN = 100;
  int n, k, c[MAXN * MAXN + 5], p[MAXN + 5], nxt[MAXN * MAXN + 5];
  pair<int, int> res[MAXN + 5];
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n * k; i++) scanf("%d", &c[i]);
    for (int L = 1; L <= n; L += k - 1) {
      int R = min(L + k - 2, n);
      vector<int> pos;
      for (int i = 1; i <= n * k; i++)
        if (L <= c[i] && c[i] <= R) pos.push_back(i);
      memset(p, 0, sizeof(p));
      memset(nxt, 0, sizeof(nxt));
      for (int i = 0; i < pos.size(); i++) {
        if (p[c[pos[i]]]) nxt[p[c[pos[i]]]] = pos[i];
        p[c[pos[i]]] = pos[i];
      }
      for (int i = L; i <= R; i++) nxt[p[i]] = n * k + 1;
      nxt[n * k + 1] = n * k + 1;
      memset(p, 0, sizeof(p));
      for (int i = 0; i < pos.size(); i++) {
        if (p[c[pos[i]]] && p[c[pos[i]]] != pos[i]) {
          res[c[pos[i]]] = make_pair(p[c[pos[i]]], pos[i]);
          p[c[pos[i]]] = 0;
          for (int j = L; j <= R; j++)
            if (p[j]) {
              while (p[j] <= pos[i]) p[j] = nxt[p[j]];
            }
        } else {
          if (!res[c[pos[i]]].first) p[c[pos[i]]] = pos[i];
        }
      }
    }
    for (int i = 1; i <= n; i++) printf("%d %d\n", res[i].first, res[i].second);
    return 0;
  }
  