description: 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>
solution: import java.util.*;
  import java.io.*;
  public class E {
  
  	public static void main(String[] args) {
  		FastScanner sc = new FastScanner();
  		StringBuilder sb = new StringBuilder();
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			int[] p = new int[n];
  			Arrays.fill(p, -1);
  			int[][] res = new int[n][2];
  			for(int i = 0; i < n; i++) {
  				Arrays.fill(res[i], -1);
  			}
  			int iter = ceil(n, k);
  			for(int i = 0; i < n*k; i++) {
  				int c = sc.nextInt()-1;
  				if(p[c] >= 0 && res[c][0] < 0) {
  					res[c][0] = p[c];
  					res[c][1] = i;
  					iter--;
  				}
  				if(iter == 0) {
  					Arrays.fill(p, -1);
  					iter = ceil(n, k);
  				}
  				else {
  					p[c] = i;
  				}
  			}
  			for(int i = 0; i < n; i++) {
  				sb.append(String.format("%d %d\n", res[i][0]+1, res[i][1]+1));
  			}
  		PrintWriter pw = new PrintWriter(System.out);
  		pw.println(sb.toString().trim());
  		pw.flush();
  	}
  	static int ceil(int n, int k) {
  		int res = n / (k-1);
  		if(n % (k-1) != 0) res++;
  		return res;
  	}
  	static class FastScanner {
  		public BufferedReader reader;
  		public StringTokenizer tokenizer;
  		public FastScanner() {
  			reader = new BufferedReader(new InputStreamReader(System.in), 32768);
  			tokenizer = null;
  		}
  		public String next() {
  			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
  				try {
  					tokenizer = new StringTokenizer(reader.readLine());
  				} catch (IOException e) {
  					throw new RuntimeException(e);
  				}
  			}
  			return tokenizer.nextToken();
  		}
  		public int nextInt() {
  			return Integer.parseInt(next());
  		}
  		public long nextLong() {
  			return Long.parseLong(next());
  		}
  		public double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		public String nextLine() {
  			try {
  				return reader.readLine();
  			} catch(IOException e) {
  				throw new RuntimeException(e);
  			}
  		}
  	}
  
  }
  