#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tt = 1;
  while (tt--) {
    long long n, k;
    cin >> n >> k;
    long long a[n * k];
    for (long long i = 0; i < n * k; i++) {
      cin >> a[i];
      a[i]--;
    }
    vector<pair<long long, long long> > intervals;
    long long last[n];
    for (long long i = 0; i < n; i++) last[i] = -1;
    for (long long i = 0; i < n * k; i++) {
      if (last[a[i]] == -1)
        last[a[i]] = i;
      else {
        intervals.push_back({i, last[a[i]]});
        last[a[i]] = i;
      }
    }
    sort(intervals.begin(), intervals.end());
    long long left[n * k];
    for (long long i = 0; i < n * k; i++) {
      left[i] = (n + k - 2) / (k - 1);
    }
    bool colour[n];
    for (long long i = 0; i < n; i++) colour[i] = false;
    pair<long long, long long> ans[n];
    long long m = intervals.size();
    for (long long i = 0; i < m; i++) {
      long long u = intervals[i].second;
      long long v = intervals[i].first;
      long long c = a[u];
      if (colour[c]) continue;
      bool ok = true;
      for (long long j = u; j <= v; j++) {
        if (left[j] == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans[c].first = u;
        ans[c].second = v;
        colour[c] = true;
        for (long long j = u; j <= v; j++) {
          left[j]--;
        }
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
    }
  }
  return 0;
}

### Prompt

Generate 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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tt = 1;
  while (tt--) {
    long long n, k;
    cin >> n >> k;
    long long a[n * k];
    for (long long i = 0; i < n * k; i++) {
      cin >> a[i];
      a[i]--;
    }
    vector<pair<long long, long long> > intervals;
    long long last[n];
    for (long long i = 0; i < n; i++) last[i] = -1;
    for (long long i = 0; i < n * k; i++) {
      if (last[a[i]] == -1)
        last[a[i]] = i;
      else {
        intervals.push_back({i, last[a[i]]});
        last[a[i]] = i;
      }
    }
    sort(intervals.begin(), intervals.end());
    long long left[n * k];
    for (long long i = 0; i < n * k; i++) {
      left[i] = (n + k - 2) / (k - 1);
    }
    bool colour[n];
    for (long long i = 0; i < n; i++) colour[i] = false;
    pair<long long, long long> ans[n];
    long long m = intervals.size();
    for (long long i = 0; i < m; i++) {
      long long u = intervals[i].second;
      long long v = intervals[i].first;
      long long c = a[u];
      if (colour[c]) continue;
      bool ok = true;
      for (long long j = u; j <= v; j++) {
        if (left[j] == 0) {
          ok = false;
          break;
        }
      }
      if (ok) {
        ans[c].first = u;
        ans[c].second = v;
        colour[c] = true;
        for (long long j = u; j <= v; j++) {
          left[j]--;
        }
      }
    }
    for (long long i = 0; i < n; i++) {
      cout << ans[i].first + 1 << " " << ans[i].second + 1 << '\n';
    }
  }
  return 0;
}

```