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>
/*
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eggyHide eggyVengeance
I need U
xiao rerun when
 */
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;

public class x1552E2
{
    public static void main(String hi[]) throws Exception
    {
        BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        int K = Integer.parseInt(st.nextToken());
        int[] arr = readArr(N*K, infile, st);
        for(int i=0; i < N*K; i++)
            arr[i]--;
        int[] freq = new int[N*K];
        int[][] res = new int[N][2];
        int[] active = new int[N];
        int BOUND = (N+K-2)/(K-1);
        for(int i=0; i < N*K; i++)
        {
            if(active[arr[i]] == 0)
                active[arr[i]] = i+1;
            else if(res[arr[i]][0] == 0)
            {
                res[arr[i]][0] = active[arr[i]];
                res[arr[i]][1] = i+1;
                int max = 0;
                for(int a=active[arr[i]]-1; a <= i; a++)
                {
                    freq[a]++;
                    max = max(max, freq[a]);
                }
                if(max == BOUND)
                    Arrays.fill(active, 0);
            }
        }
        StringBuilder sb = new StringBuilder();
        for(int c=0; c < N; c++)
            sb.append(res[c][0]+" "+res[c][1]+"\n");
        System.out.print(sb);
    }
    public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
    {
        int[] arr = new int[N];
        st = new StringTokenizer(infile.readLine());
        for(int i=0; i < N; i++)
            arr[i] = Integer.parseInt(st.nextToken());
        return arr;
    }
}
/*
ai and bi of chosen intervals must be consecutive occurrences of that color

3 4
1 3 2 3 2 1 3 1 3 2 1 2

3 4
1 3 2 3 2 1 3 1 2 1 2 3

6 3
1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6
 */