#include <bits/stdc++.h>
using namespace std;
void ProGamerMove() {
  int n, k;
  cin >> n >> k;
  vector<int> f(n * k), lst(n, -1), tmp((n + k - 2) / (k - 1)), done(n);
  vector<pair<int, int>> res(n);
  multiset<int> can(tmp.begin(), tmp.end());
  int i = 0;
  for (int& v : f) {
    cin >> v;
    --v;
    if (~lst[v] && !done[v]) {
      int l = lst[v], r = i + 1;
      if (*can.begin() <= l) {
        res[v] = {l, r};
        can.erase(prev(can.lower_bound(l + 1)));
        can.insert(r);
        done[v] = true;
      }
    }
    lst[v] = i;
    ++i;
  }
  for (auto [l, r] : res) cout << l + 1 << " " << r << "\n";
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int tc = 1;
  while (tc--) ProGamerMove();
}

### 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;
void ProGamerMove() {
  int n, k;
  cin >> n >> k;
  vector<int> f(n * k), lst(n, -1), tmp((n + k - 2) / (k - 1)), done(n);
  vector<pair<int, int>> res(n);
  multiset<int> can(tmp.begin(), tmp.end());
  int i = 0;
  for (int& v : f) {
    cin >> v;
    --v;
    if (~lst[v] && !done[v]) {
      int l = lst[v], r = i + 1;
      if (*can.begin() <= l) {
        res[v] = {l, r};
        can.erase(prev(can.lower_bound(l + 1)));
        can.insert(r);
        done[v] = true;
      }
    }
    lst[v] = i;
    ++i;
  }
  for (auto [l, r] : res) cout << l + 1 << " " << r << "\n";
}
signed main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  int tc = 1;
  while (tc--) ProGamerMove();
}

```