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.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws Exception {
        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());
        st = new StringTokenizer(br.readLine());
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(st.nextToken());
        }

        int x = 0;
        int y = -1;
        int bestX = -1;
        int bestY = -2;
        int[] b = new int[1000002];
        int curr = 0;
        while (true) {
            y++;
            b[a[y]]++;
            if (b[a[y]] == 1) {
                curr++;
            }
            while (curr > k) {
                b[a[x]]--;
                if (b[a[x]] == 0) {
                    curr--;
                }
                x++;
            }
            if (y - x > bestY - bestX) {
                bestX = x;
                bestY = y;
            }
            if (y == n - 1) {
                break;
            }
        }
        System.out.println((bestX + 1) + " " + (bestY + 1));
    }
}