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.io.*;
import java.util.*;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();
		int tcCnt = 1;
		for (int tc = 1; tc <= tcCnt; tc++)
			solver.solve(tc, in, out);
		out.close();
	}

	static class Task {
		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int k = in.nextInt();
			int m = (n+k-2)/(k-1);
			int[][] pos = new int[n][k];
			int[] ptr = new int[n];
			for (int i = 0; i < n*k; i++) {
				int c = in.nextInt() - 1;
				pos[c][ptr[c]++] = i;
			}
			Integer[] colors = new Integer[n];
			for (int i = 0; i < n; i++) {
				colors[i] = i;
			}
			int[][] intervals = new int[n][2];
			int cnt = 0;
			for (int i = 0; cnt < n; i++) {
				int idx = i;
				Arrays.sort(colors, Comparator.comparingInt(p -> pos[p][idx + 1]));
				for (int j = 0, tmpcnt = 0; j < n && tmpcnt < m; j++) {
					int col = colors[j];
					if (intervals[col][0] != 0) {
						continue;
					}
					tmpcnt++;
					cnt++;
					intervals[col][0] = pos[col][i] + 1;
					intervals[col][1] = pos[col][i+1] + 1;
				}
			}
			for (int i = 0; i < n; i++) {
				out.println(intervals[i][0]+" "+intervals[i][1]);
			}
		}

	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}

		public String nextToken() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				tokenizer = new StringTokenizer(nextLine());
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(nextToken());
		}

		public double nextDouble() {
			return Double.parseDouble(nextToken());
		}

		public long nextLong() {
			return Long.parseLong(nextToken());
		}

		public int[] nextIntArray(int size) {
			int[] arr = new int[size];
			for (int i = 0; i < size; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}

	}
}