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.util.*;
import java.lang.*;
import java.io.*;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String line;
        String[] ar;
        StringBuilder sb = new StringBuilder();
        //-----------------------------------------------------------------------
        ar = br.readLine().split(" ");
        int n = Integer.parseInt(ar[0]);
        int k = Integer.parseInt(ar[1]);
        int[] a = new int[n];
        ar = br.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(ar[i]);
        }
        int i = 0, j = 0, max = 0, l = 0, r = 0, ck = 0;
        int[] f = new int[1000001];
        while (j < n) {
            if (f[a[j]] > 0) {
                f[a[j]]++;
                j++;
            } else {
                ck++;
                f[a[j]]++;
                if (ck > k) {
                    while (i <= j) {
                        f[a[i]]--;
                        if (f[a[i]] == 0) {
                            ck--;
                            i++;
                            break;
                        }
                        i++;
                    }
                }
                j++;
            }
            if (j - i > max) {
                max = j - i;
                r = j; l = i;
            }
        }
        sb.append((l + 1) + " " + (r));
        //-----------------------------------------------------------------------
        System.out.print(sb);
    }
}