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;
vector<int> c[102];
int l[102], r[102], U[102];
int idx[102];
int n, k;
void choose(int s, int e) {
  int now = 0;
  for (int t = 1; t <= e - s + 1; t++) {
    int opt = 100000, w = -1, x = 0;
    for (int i = s; i <= e; i++) {
      if (U[i]) continue;
      int cnt = 0;
      for (int j = 0; j < k; j++) {
        if (c[i][j] > now) cnt++;
        if (cnt == 2) {
          if (opt > c[i][j]) {
            opt = c[i][j];
            x = j;
            w = i;
          }
          break;
        }
      }
    }
    U[w] = true;
    now = opt;
    l[w] = c[w][x - 1];
    r[w] = c[w][x];
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i <= n * k; i++) {
    int C;
    cin >> C;
    c[C].push_back(i);
  }
  for (int i = 1; i <= n; i++) U[i] = 0;
  for (int i = 1; i <= n; i = i + k - 1) {
    choose(i, min(i + k - 2, n));
  }
  for (int i = 1; i <= n; i++) cout << l[i] << " " << r[i] << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) solve();
}
