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
//package que_a;

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

public class utkarsh {

    InputStream is;
    PrintWriter out;
    
    long mod = (long) (1e9 + 7), inf = (long) (3e18);
    
    void solve() {
        int n = ni(), k = ni();
        int a[] = na(n);
        int feq[] = new int[1000100];
        int l = 0, r = 0, cnt = 0, x = 0, y = 0;
        while(r < n) {
            if(feq[ a[r] ] == 0) cnt++;
            feq[ a[r] ]++;
            while(cnt > k) {
                feq[ a[l] ]--;
                if(feq[ a[l] ] == 0)    cnt--;
                l++;
            }
            if((r-l) > (y-x)) {
                x = l;  y = r;
            }
            r++;
        }
        out.println((x+1) +" "+ (y+1));
    }
    
    long mp(long b, long e) {
        long r = 1;
        while(e > 0) {
            if((e&1) == 1)  r = (r * b) % mod;
            b = (b * b) % mod;
            e >>= 1;
        }
        return r;
    }

    //---------- I/O Template ----------
    
    public static void main(String[] args) { new utkarsh().run(); }
    void run() { 
        is = System.in;
        out = new PrintWriter(System.out);
        solve();
        out.flush();
    }
    
    byte input[] = new byte[1024];
    int len = 0, ptr = 0;
    
    int readByte() { 
        if(ptr >= len) { ptr = 0; 
            try { len = is.read(input); } 
            catch(IOException e) { throw new InputMismatchException(); } 
            if(len <= 0) { return -1; } 
        } return input[ptr++];
    }
    boolean isSpaceChar(int c) { return !( c >= 33 && c <= 126 ); }
    int skip() { 
        int b = readByte(); 
        while(b != -1 && isSpaceChar(b)) { b = readByte(); } 
        return b;
    }
    
    char nc() { return (char)skip(); }
    String ns() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while(!isSpaceChar(b)) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    String nLine() { 
        int b = skip(); 
        StringBuilder sb = new StringBuilder(); 
        while( !(isSpaceChar(b) && b != ' ') ) { sb.appendCodePoint(b); b = readByte(); } 
        return sb.toString();
    }
    int ni() { 
        int n = 0, b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        if(b == -1) { return -1; }  //no input 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }
    long nl() { 
        long n = 0L;    int b = readByte(); 
        boolean minus = false; 
        while(b != -1 && !( (b >= '0' && b <= '9') || b == '-')) { b = readByte(); } 
        if(b == '-') { minus = true; b = readByte(); } 
        while(b >= '0' && b <= '9') { n = n * 10 + (b - '0'); b = readByte(); } 
        return minus ? -n : n;
    }

    double nd() { return Double.parseDouble(ns()); }
    float nf() { return Float.parseFloat(ns()); }
    int[] na(int n) { 
        int a[] = new int[n]; 
        for(int i = 0; i < n; i++) { a[i] = ni(); } 
        return a;
    }
    char[] ns(int n) { 
        char c[] = new char[n]; 
        int i, b = skip(); 
        for(i = 0; i < n; i++) { 
            if(isSpaceChar(b)) { break; } 
            c[i] = (char)b; b = readByte(); 
        } return i == n ? c : Arrays.copyOf(c,i);
    }
}