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>


import java.util.Scanner;

public class E {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);

        solve(s.nextInt(), s.nextInt(), s);
    }


    public static void solve(int c, int r, Scanner s) {
        int v = c / (r - 1) + (c % (r - 1) == 0 ? 0 : 1);
        int e = c / v + (c % v == 0 ? 0 : 1);

        int a[] = new int[c * r];
        for (int i = 0; i < a.length; i++) {
            a[i] = s.nextInt() - 1;
        }
        
        boolean done[];
        boolean visited[];
        int first[];
        int l;
        int min;
        int index;
        int t;

        String ans[] = new String[c];

        for (int i = 0; i < v; i++) {
            min = i * e;
            if (i != v - 1) {
                l = e;
            } else {
                l = c % e;
                if (l == 0) l = e;
            }

            done = new boolean[l];
            index = 0;
            for (int j = 0; j < l; j++) {
                first = new int[l];
                visited = new boolean[l];
                for (; index < a.length; index++) {
                    t = a[index];
                    if (t >= min && t < min + l) {
                        if (!done[t - min]) {
                            if (visited[t - min]) {
                                ans[t] = (first[t - min] + 1) + " " + (index + 1);
                                done[t - min] = true;
                                break;
                            } else {
                                visited[t - min] = true;
                                first[t - min] = index;
                            }
                        }
                    }
                }
            }
        }

        for (int i = 0; i < c; i++) {
            System.out.println(ans[i]);
        }
    }
}
