The array a with n integers is given. Let's call the sequence of one or more consecutive elements in a segment. Also let's call the segment k-good if it contains no more than k different values.

Find any longest k-good segment.

As the input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printf instead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

Input

The first line contains two integers n, k (1 ≤ k ≤ n ≤ 5·105) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 106) — the elements of the array a.

Output

Print two integers l, r (1 ≤ l ≤ r ≤ n) — the index of the left and the index of the right ends of some k-good longest segment. If there are several longest segments you can print any of them. The elements in a are numbered from 1 to n from left to right.

Examples

Input

5 5
1 2 3 4 5


Output

1 5


Input

9 3
6 5 1 2 3 2 1 4 5


Output

3 7


Input

3 1
1 2 3


Output

1 1
import java.io.*;
import java.util.*;
import java.lang.Math;
public class longestKGoodSegment {
  public static void main (String [] args) throws IOException {
    
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        StringTokenizer s = new StringTokenizer(f.readLine());
        int[] hs = new int[1000009];
        int distinct = 0;
        int n = Integer.parseInt(s.nextToken());
        int k = Integer.parseInt(s.nextToken());
        int[] array = new int[n + 1];
        s = new StringTokenizer(f.readLine());
        for(int i = 1; i < n + 1; i++){
            array[i] = Integer.parseInt(s.nextToken());
        }
        int answer = 0;
        int pointer1 = 1;
        int pointer2 = 1;
        int[] longest = {0, 0, 0};
        while(pointer1 < array.length && pointer2 <= array.length){
            
            while(pointer2 < array.length && verify(hs, distinct, k, array[pointer2])){
                pointer2++;
                if(hs[array[pointer2 - 1]] == 0){
                    distinct++;
                }
                hs[array[pointer2 - 1]]++;
            }
            
            if(pointer2 - pointer1 > longest[0]){
                longest[0] = pointer2 - pointer1;
                longest[1] = pointer1;
                longest[2] = pointer2;
            }
            
            if(pointer2 == array.length){
                break;
            }
            hs[array[pointer1]] -= 1;
            if(hs[array[pointer1]] == 0){
                distinct -= 1;
            }
            
            hs[array[pointer2]] += 1;
            if(hs[array[pointer2]] == 1){
                distinct += 1;
            }
            pointer1++;
            pointer2++;
        }
        out.println(longest[1] + " " + (longest[2] - 1));
        out.close();
  }
  public static boolean verify(int[] hs, int distinct, int k, int num){
      int ddd = distinct;
      if(hs[num] == 0){
          ddd++;
      }
      if(ddd <= k){
          return true;
      }
      return false;
  }
}
