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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main
{

    void solve() throws Exception
    {
        int n = nextInt();
        TreeSet<Integer> s = new TreeSet<>();
        ArrayList<Integer> ans = new ArrayList<>();
        for( int i = 0; i < n; i++ )
        {
            int a = nextInt();
            if( s.contains( a ) )
            {
                ans.add( i );
                s.clear();
            }
            else
                s.add( a );
        }
        if( ans.size() == 0 )
        {
            out.println( -1 );
        }
        else
        {
            out.println( ans.size() );
            int bg = 0;
            for( int i = 0; i < ans.size() - 1; i++ )
            {
                out.println( ( bg + 1 ) + " " + ( ans.get( i ) + 1 ) );
                bg = ans.get( i ) + 1;
            }
            out.println( ( bg + 1 ) + " " + n );
        }
    }

    public static void main( String[] args )
    {
        new Main().run();
    }

    public void run()
    {
        try
        {
            in = new BufferedReader( new InputStreamReader( System.in ) );
            out = new PrintWriter( System.out );
            solve();
        }
        catch( Exception e )
        {
            e.printStackTrace();
            System.exit( 1 );
        }
        finally
        {
            out.close();
        }
    }

    BufferedReader in;
    PrintWriter out;
    StringTokenizer st;

    String nextToken() throws IOException
    {
        while( st == null || !st.hasMoreTokens() ) st = new StringTokenizer( in.readLine() );
        return st.nextToken();
    }

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

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

}