#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
int main() {
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  };
  int n, k;
  cin >> n >> k;
  vector<int> v(n * k);
  for (int i = 0; i < (n * k); i++) {
    cin >> v[i];
    v[i]--;
  }
  vector<vector<int>> g(n);
  vector<vector<pair<int, int>>> pos(k);
  vector<bool> used(n);
  vector<pair<int, int>> ans(n);
  for (int i = 0; i < (n * k); i++) g[v[i]].push_back(i);
  for (int i = 0; i < (k); i++) {
    for (int j = 0; j < (n); j++) pos[i].push_back({g[j][i], j});
  }
  int ost = n, step = 1;
  while (ost > 0) {
    sort(pos[step].begin(), pos[step].end());
    int l = 0;
    int f = (n + k - 2) / (k - 1);
    while (ost > 0 && l < n && f > 0) {
      int col = pos[step][l].second;
      if (used[col]) {
        l++;
        continue;
      }
      used[col] = 1;
      ans[col] = {g[col][step - 1], g[col][step]};
      f--;
      ost--;
      l++;
    }
    step++;
  }
  for (auto u : ans) cout << u.first + 1 << " " << u.second + 1 << endl;
}

### Prompt

Your challenge is to write 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;
const long long MOD = 998244353;
int main() {
  {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  };
  int n, k;
  cin >> n >> k;
  vector<int> v(n * k);
  for (int i = 0; i < (n * k); i++) {
    cin >> v[i];
    v[i]--;
  }
  vector<vector<int>> g(n);
  vector<vector<pair<int, int>>> pos(k);
  vector<bool> used(n);
  vector<pair<int, int>> ans(n);
  for (int i = 0; i < (n * k); i++) g[v[i]].push_back(i);
  for (int i = 0; i < (k); i++) {
    for (int j = 0; j < (n); j++) pos[i].push_back({g[j][i], j});
  }
  int ost = n, step = 1;
  while (ost > 0) {
    sort(pos[step].begin(), pos[step].end());
    int l = 0;
    int f = (n + k - 2) / (k - 1);
    while (ost > 0 && l < n && f > 0) {
      int col = pos[step][l].second;
      if (used[col]) {
        l++;
        continue;
      }
      used[col] = 1;
      ans[col] = {g[col][step - 1], g[col][step]};
      f--;
      ost--;
      l++;
    }
    step++;
  }
  for (auto u : ans) cout << u.first + 1 << " " << u.second + 1 << endl;
}

```