#include <bits/stdc++.h>
using namespace std;
vector<array<int, 2>> solve_case(int balls, vector<int> &arr) {
  vector<bool> alive(balls - 1, true);
  vector<array<int, 2>> answer(balls - 1);
  int cur = 0;
  for (int i = 0; i < balls - 1; i++) {
    map<int, vector<int>> cnt;
    int found = -1;
    while (true) {
      if (alive[arr[cur]]) {
        cnt[arr[cur]].push_back(cur);
        if (cnt[arr[cur]].size() > 1) {
          found = arr[cur];
          break;
        }
      }
      cur += 1;
    }
    assert(found + 1);
    answer[found][0] = cnt[found][0];
    answer[found][1] = cnt[found][1];
    alive[found] = false;
    cur += 1;
  }
  return answer;
}
void solve_single(int balls, vector<array<int, 2>> &arr, int offset,
                  vector<array<int, 2>> &answer) {
  vector<int> single(balls * (balls - 1));
  for (int i = 0; i < balls * (balls - 1); i++) {
    single[i] = arr[i][0];
  }
  vector<array<int, 2>> ans = solve_case(balls, single);
  for (int i = 0; i < balls - 1; i++) {
    answer[offset + i][0] = arr[ans[i][0]][1];
    answer[offset + i][1] = arr[ans[i][1]][1];
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int n_orig = n;
  vector<int> arr(n * k);
  for (int i = 0; i < n * k; i++) {
    scanf("%d", &arr[i]);
    arr[i] -= 1;
  }
  while (n % (k - 1) != 0) {
    for (int i = 0; i < k; i++) {
      arr.push_back(n);
    }
    n += 1;
  }
  int step = k - 1;
  vector<array<int, 2>> answer(n);
  for (int col = 0; col < n; col += step) {
    int col_start = col, col_end = col + step;
    vector<array<int, 2>> solve;
    for (int i = 0; i < n * k; i++) {
      if (col_start <= arr[i] && arr[i] < col_end) {
        solve.push_back({arr[i] - col_start, i});
      }
    }
    assert((int)solve.size() == k * (k - 1));
    solve_single(k, solve, col_start, answer);
  }
  for (int i = 0; i < n_orig; i++) {
    printf("%d %d\n", answer[i][0] + 1, answer[i][1] + 1);
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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>
using namespace std;
vector<array<int, 2>> solve_case(int balls, vector<int> &arr) {
  vector<bool> alive(balls - 1, true);
  vector<array<int, 2>> answer(balls - 1);
  int cur = 0;
  for (int i = 0; i < balls - 1; i++) {
    map<int, vector<int>> cnt;
    int found = -1;
    while (true) {
      if (alive[arr[cur]]) {
        cnt[arr[cur]].push_back(cur);
        if (cnt[arr[cur]].size() > 1) {
          found = arr[cur];
          break;
        }
      }
      cur += 1;
    }
    assert(found + 1);
    answer[found][0] = cnt[found][0];
    answer[found][1] = cnt[found][1];
    alive[found] = false;
    cur += 1;
  }
  return answer;
}
void solve_single(int balls, vector<array<int, 2>> &arr, int offset,
                  vector<array<int, 2>> &answer) {
  vector<int> single(balls * (balls - 1));
  for (int i = 0; i < balls * (balls - 1); i++) {
    single[i] = arr[i][0];
  }
  vector<array<int, 2>> ans = solve_case(balls, single);
  for (int i = 0; i < balls - 1; i++) {
    answer[offset + i][0] = arr[ans[i][0]][1];
    answer[offset + i][1] = arr[ans[i][1]][1];
  }
}
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  int n_orig = n;
  vector<int> arr(n * k);
  for (int i = 0; i < n * k; i++) {
    scanf("%d", &arr[i]);
    arr[i] -= 1;
  }
  while (n % (k - 1) != 0) {
    for (int i = 0; i < k; i++) {
      arr.push_back(n);
    }
    n += 1;
  }
  int step = k - 1;
  vector<array<int, 2>> answer(n);
  for (int col = 0; col < n; col += step) {
    int col_start = col, col_end = col + step;
    vector<array<int, 2>> solve;
    for (int i = 0; i < n * k; i++) {
      if (col_start <= arr[i] && arr[i] < col_end) {
        solve.push_back({arr[i] - col_start, i});
      }
    }
    assert((int)solve.size() == k * (k - 1));
    solve_single(k, solve, col_start, answer);
  }
  for (int i = 0; i < n_orig; i++) {
    printf("%d %d\n", answer[i][0] + 1, answer[i][1] + 1);
  }
  return 0;
}

```