#include <bits/stdc++.h>
void Solve() {
  int n = 0, k = 0;
  std::cin >> n >> k;
  std::vector<std::vector<int>> by_col(n);
  int c = 0;
  for (int i = 0; i < n * k; ++i) {
    std::cin >> c;
    by_col[c - 1].push_back(i);
  }
  std::vector<std::pair<int, int>> ans(n);
  for (int i = 0; i < n; i += k - 1) {
    std::vector<int> ptr(k - 1, 0);
    int last_right_end = 0;
    for (int round = 0; round < std::min(k - 1, n - i); ++round) {
      int min_right_end = n * k + 1;
      int argmax = -1;
      for (int j = 0; j < std::min(k - 1, n - i); ++j) {
        if (ptr[j] != -1) {
          while (by_col[i + j][ptr[j]] < last_right_end) {
            ++ptr[j];
          }
          if (min_right_end > by_col[i + j][ptr[j] + 1]) {
            min_right_end = by_col[i + j][ptr[j] + 1];
            argmax = j;
          }
        }
      }
      last_right_end = min_right_end;
      ans[i + argmax] = {by_col[i + argmax][ptr[argmax]],
                         by_col[i + argmax][ptr[argmax] + 1]};
      ptr[argmax] = -1;
    }
  }
  for (auto& interval : ans) {
    std::cout << interval.first + 1 << ' ' << interval.second + 1 << '\n';
  }
}
int main() { Solve(); }

### Prompt

Construct a CPP code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
void Solve() {
  int n = 0, k = 0;
  std::cin >> n >> k;
  std::vector<std::vector<int>> by_col(n);
  int c = 0;
  for (int i = 0; i < n * k; ++i) {
    std::cin >> c;
    by_col[c - 1].push_back(i);
  }
  std::vector<std::pair<int, int>> ans(n);
  for (int i = 0; i < n; i += k - 1) {
    std::vector<int> ptr(k - 1, 0);
    int last_right_end = 0;
    for (int round = 0; round < std::min(k - 1, n - i); ++round) {
      int min_right_end = n * k + 1;
      int argmax = -1;
      for (int j = 0; j < std::min(k - 1, n - i); ++j) {
        if (ptr[j] != -1) {
          while (by_col[i + j][ptr[j]] < last_right_end) {
            ++ptr[j];
          }
          if (min_right_end > by_col[i + j][ptr[j] + 1]) {
            min_right_end = by_col[i + j][ptr[j] + 1];
            argmax = j;
          }
        }
      }
      last_right_end = min_right_end;
      ans[i + argmax] = {by_col[i + argmax][ptr[argmax]],
                         by_col[i + argmax][ptr[argmax] + 1]};
      ptr[argmax] = -1;
    }
  }
  for (auto& interval : ans) {
    std::cout << interval.first + 1 << ' ' << interval.second + 1 << '\n';
  }
}
int main() { Solve(); }

```