Problem: 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

Solution (in JAVA):

Solution:
import java.io.*; import java.util.*;
public class CF0616D {

    /**
     * 2 pointers
     */
    static int MAXN=1000001 ;
    public static void main(String[] args) throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int N=Integer.parseInt(st.nextToken());
        int K=Integer.parseInt(st.nextToken());
        int[] a=new int[N]; int[] prefix=new int[MAXN];
        st=new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            a[i]=Integer.parseInt(st.nextToken());
        }
        int cur=0; int best=1; int bestl=0;
        int r=0; 
        for (int l = 0; l < N; l++) {
            if(l>0){
                prefix[a[l-1]]--; 
                if(prefix[a[l-1]]==0){
                    cur--;
                }
            }
            //System.out.println(l+" "+r+" "+cur);
            while(r<N && cur<=K){
                
                prefix[a[r]]++;
                if(prefix[a[r]]==1){
                    cur++;
                }
                r++;
                //System.out.println(l+" "+r+" "+cur);
            }
            
            if(cur<=K){
                if(r-l>best){
                    best=r-l; bestl=l;
                }
            }else if(r-l-1>best){//cur-K can be -1 or 0
                //If 0 then include, if -1 then exclude r
                best=r-l-1; bestl=l;
            }
        }
        System.out.println(bestl+1+" "+(bestl+best));
    }
    
}