#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pi = pair<int, int>;
using vi = vector<int>;
const int nax = 110;
int n, k;
int cnt[nax];
bool done[nax];
int last[nax], last2[nax];
pi ans[nax];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int op = 0;
  int l = (n - 1) / (k - 1) + 1;
  vi used = {};
  for (int i = 0; i < n * k; ++i) {
    int x;
    cin >> x;
    cnt[x]++;
    last2[x] = last[x];
    last[x] = i + 1;
    if (cnt[x] == 2 && !done[x]) {
      op++;
      used.push_back(x);
    }
    if (op == l || i == n * k - 1) {
      op = 0;
      for (int y : used) {
        ans[y] = {last2[y], last[y]};
        done[y] = true;
      }
      used.clear();
      for (int j = 1; j <= n; ++j) {
        cnt[j] = 0;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    auto x = ans[i];
    cout << x.first << " " << x.second << "\n";
  }
}

### Prompt

Create a solution in cpp for 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;
using ll = long long;
using pi = pair<int, int>;
using vi = vector<int>;
const int nax = 110;
int n, k;
int cnt[nax];
bool done[nax];
int last[nax], last2[nax];
pi ans[nax];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int op = 0;
  int l = (n - 1) / (k - 1) + 1;
  vi used = {};
  for (int i = 0; i < n * k; ++i) {
    int x;
    cin >> x;
    cnt[x]++;
    last2[x] = last[x];
    last[x] = i + 1;
    if (cnt[x] == 2 && !done[x]) {
      op++;
      used.push_back(x);
    }
    if (op == l || i == n * k - 1) {
      op = 0;
      for (int y : used) {
        ans[y] = {last2[y], last[y]};
        done[y] = true;
      }
      used.clear();
      for (int j = 1; j <= n; ++j) {
        cnt[j] = 0;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    auto x = ans[i];
    cout << x.first << " " << x.second << "\n";
  }
}

```