#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> c(n * k + 1);
  for (long long z = 1; (1 > 0 and z < n * k + 1) or (1 < 0 and z > n * k + 1);
       z += 1)
    cin >> c[z];
  ;
  vector<pair<long long, long long>> ivl(n + 1, {-1, -1});
  vector<pair<vector<long long>, long long>> a(n, {vector<long long>(), 0});
  for (long long i = 1; (1 > 0 and i < n * k + 1) or (1 < 0 and i > n * k + 1);
       i += 1)
    a[c[i] - 1].first.push_back(i);
  for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
    a[i].second = i + 1;
  long long t = 1;
  function<bool(pair<vector<long long>, long long>,
                pair<vector<long long>, long long>)>
      comp = [&](pair<vector<long long>, long long> a1,
                 pair<vector<long long>, long long> a2) {
        return a1.first[t] < a2.first[t];
      };
  for (; t < k; ++t) {
    long long m = a.size();
    sort(a.begin(), a.end(), comp);
    for (long long i = 0;
         (1 > 0 and i < min(m, n / (k - 1) + (n % (k - 1) > 0))) or
         (1 < 0 and i > min(m, n / (k - 1) + (n % (k - 1) > 0)));
         i += 1)
      ivl[a[i].second] = {a[i].first[t - 1], a[i].first[t]};
    if (m > n / (k - 1) + (n % (k - 1) > 0)) {
      vector<pair<vector<long long>, long long>> b;
      for (long long i = n / (k - 1) + (n % (k - 1) > 0);
           (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
        b.push_back(a[i]);
      a = b;
    } else
      break;
  }
  for (long long i = 1; (1 > 0 and i < n + 1) or (1 < 0 and i > n + 1); i += 1)
    cout << ivl[i].first << " " << ivl[i].second << endl;
  return 0;
}

### 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  vector<long long> c(n * k + 1);
  for (long long z = 1; (1 > 0 and z < n * k + 1) or (1 < 0 and z > n * k + 1);
       z += 1)
    cin >> c[z];
  ;
  vector<pair<long long, long long>> ivl(n + 1, {-1, -1});
  vector<pair<vector<long long>, long long>> a(n, {vector<long long>(), 0});
  for (long long i = 1; (1 > 0 and i < n * k + 1) or (1 < 0 and i > n * k + 1);
       i += 1)
    a[c[i] - 1].first.push_back(i);
  for (long long i = 0; (1 > 0 and i < n) or (1 < 0 and i > n); i += 1)
    a[i].second = i + 1;
  long long t = 1;
  function<bool(pair<vector<long long>, long long>,
                pair<vector<long long>, long long>)>
      comp = [&](pair<vector<long long>, long long> a1,
                 pair<vector<long long>, long long> a2) {
        return a1.first[t] < a2.first[t];
      };
  for (; t < k; ++t) {
    long long m = a.size();
    sort(a.begin(), a.end(), comp);
    for (long long i = 0;
         (1 > 0 and i < min(m, n / (k - 1) + (n % (k - 1) > 0))) or
         (1 < 0 and i > min(m, n / (k - 1) + (n % (k - 1) > 0)));
         i += 1)
      ivl[a[i].second] = {a[i].first[t - 1], a[i].first[t]};
    if (m > n / (k - 1) + (n % (k - 1) > 0)) {
      vector<pair<vector<long long>, long long>> b;
      for (long long i = n / (k - 1) + (n % (k - 1) > 0);
           (1 > 0 and i < m) or (1 < 0 and i > m); i += 1)
        b.push_back(a[i]);
      a = b;
    } else
      break;
  }
  for (long long i = 1; (1 > 0 and i < n + 1) or (1 < 0 and i > n + 1); i += 1)
    cout << ivl[i].first << " " << ivl[i].second << endl;
  return 0;
}

```