#include <bits/stdc++.h>
using namespace std;
int N, K, arr[10005], cnt[105], pos[105][105];
vector<pair<int, int>> ans;
int main() {
  cin >> N >> K;
  int sz = (N - 1) / (K - 1) + 1;
  vector<int> v;
  for (int i = 1; i <= N; ++i) {
    v.push_back(i);
  }
  for (int i = 1; i <= N * K; ++i) {
    cin >> arr[i];
    pos[arr[i]][cnt[arr[i]]++] = i;
  }
  for (int i = 1; v.size(); ++i) {
    sort(v.begin(), v.end(),
         [&](int a, int b) { return pos[a][i] > pos[b][i]; });
    for (int j = 0; j < sz && v.size(); ++j) {
      int cur = v.back();
      v.pop_back();
      ans.push_back({pos[cur][i - 1], pos[cur][i]});
    }
  }
  sort(ans.begin(), ans.end(), [&](pair<int, int> a, pair<int, int> b) {
    return arr[a.first] < arr[b.first];
  });
  for (auto [a, b] : ans) {
    cout << a << " " << b << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
int N, K, arr[10005], cnt[105], pos[105][105];
vector<pair<int, int>> ans;
int main() {
  cin >> N >> K;
  int sz = (N - 1) / (K - 1) + 1;
  vector<int> v;
  for (int i = 1; i <= N; ++i) {
    v.push_back(i);
  }
  for (int i = 1; i <= N * K; ++i) {
    cin >> arr[i];
    pos[arr[i]][cnt[arr[i]]++] = i;
  }
  for (int i = 1; v.size(); ++i) {
    sort(v.begin(), v.end(),
         [&](int a, int b) { return pos[a][i] > pos[b][i]; });
    for (int j = 0; j < sz && v.size(); ++j) {
      int cur = v.back();
      v.pop_back();
      ans.push_back({pos[cur][i - 1], pos[cur][i]});
    }
  }
  sort(ans.begin(), ans.end(), [&](pair<int, int> a, pair<int, int> b) {
    return arr[a.first] < arr[b.first];
  });
  for (auto [a, b] : ans) {
    cout << a << " " << b << endl;
  }
  return 0;
}

```