#include <bits/stdc++.h>
using namespace std;
constexpr long long MAXN = 100;
long long arr[MAXN * MAXN];
long long handles[MAXN][MAXN];
pair<long long, long long> ans[MAXN];
int32_t main() {
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  long long handles_ct[MAXN];
  fill_n(handles_ct, n, 0);
  for (long long i = 0; i < n * k; ++i) {
    cin >> arr[i];
    --arr[i];
    handles[arr[i]][handles_ct[arr[i]]++] = i;
  }
  fill_n(handles_ct, n, 1);
  long long magic = (n + (k - 2)) / (k - 1);
  for (long long i = 0; i < n;) {
    long long cutoff = -1;
    for (long long ii = 0; i < n && ii < magic; ++ii, ++i) {
      long long best = -1;
      for (long long j = 0; j < n; ++j) {
        if (ans[j].second == 0) {
          if (best == -1 ||
              handles[j][handles_ct[j]] < handles[best][handles_ct[best]]) {
            best = j;
          }
        }
      }
      ans[best].first = handles[best][handles_ct[best] - 1];
      cutoff = ans[best].second = handles[best][handles_ct[best]];
    }
    if (cutoff == -1) break;
    for (long long j = 0; j < n; ++j) {
      if (ans[j].second == 0) {
        if (handles[j][handles_ct[j] - 1] <= cutoff) ++handles_ct[j];
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << '\n';
  }
}

### 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>
using namespace std;
constexpr long long MAXN = 100;
long long arr[MAXN * MAXN];
long long handles[MAXN][MAXN];
pair<long long, long long> ans[MAXN];
int32_t main() {
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  long long handles_ct[MAXN];
  fill_n(handles_ct, n, 0);
  for (long long i = 0; i < n * k; ++i) {
    cin >> arr[i];
    --arr[i];
    handles[arr[i]][handles_ct[arr[i]]++] = i;
  }
  fill_n(handles_ct, n, 1);
  long long magic = (n + (k - 2)) / (k - 1);
  for (long long i = 0; i < n;) {
    long long cutoff = -1;
    for (long long ii = 0; i < n && ii < magic; ++ii, ++i) {
      long long best = -1;
      for (long long j = 0; j < n; ++j) {
        if (ans[j].second == 0) {
          if (best == -1 ||
              handles[j][handles_ct[j]] < handles[best][handles_ct[best]]) {
            best = j;
          }
        }
      }
      ans[best].first = handles[best][handles_ct[best] - 1];
      cutoff = ans[best].second = handles[best][handles_ct[best]];
    }
    if (cutoff == -1) break;
    for (long long j = 0; j < n; ++j) {
      if (ans[j].second == 0) {
        if (handles[j][handles_ct[j] - 1] <= cutoff) ++handles_ct[j];
      }
    }
  }
  for (long long i = 0; i < n; ++i) {
    cout << ans[i].first + 1 << ' ' << ans[i].second + 1 << '\n';
  }
}

```