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>
#include <bits/stdc++.h>
using namespace std;
int const N = 1e5 + 123;
void TestCase() {
  int n, k;
  cin >> n >> k;
  vector<int> v(n * k);
  for (auto& x : v) cin >> x;
  vector<int> cnt(n * k + 5, 0);
  vector<int> pre(n * k + 5, -1);
  vector<int> last(n * k + 5, -1);
  vector<pair<int, pair<int, int>>> ans;
  int limit = ((n % (k - 1) == 0) ? n / (k - 1) + 1 : n / (k - 1) + 2);
  for (int i = 0; i < v.size(); i++) {
    if (pre[v[i]] != -1 && last[v[i]] == -1) {
      int L = pre[v[i]];
      int R = i;
      bool ok = true;
      for (int j = L; j <= R; j++) {
        if (cnt[j] == limit - 1) ok = false;
      }
      if (ok) {
        for (int j = L; j <= R; j++) {
          cnt[j]++;
        }
        last[v[i]] = i;
        ans.push_back({v[i], {L + 1, R + 1}});
      }
    }
    pre[v[i]] = i;
  }
  sort(ans.begin(), ans.end());
  for (auto x : ans) {
    cout << x.second.first << ' ' << x.second.second << '\n';
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  TestCase();
}
