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.Scanner;

/**
 * Created by fleeon on 1/13/16.
 */
public class P616D {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int K = scanner.nextInt();
        int ni[] = new int[N + 1];
        int e[] = new int[(int) 1e6 + 1];
        int cnt = 0, s = 0, maxS = -1, maxL = 0;
        for (int n = 0; n < N; n++) {
            ni[n] = scanner.nextInt();
            if (e[ni[n]] == 0) cnt++;
            e[ni[n]]++;
            if (cnt > K) {
                e[ni[s]]--;
                if (e[ni[s++]] == 0) cnt--;
            }
            if (cnt <= K) {
                if (maxL < n - s + 1) {
                    maxL = n - s + 1;
                    maxS = s;
                }
            }
        }
        System.out.println((maxS + 1) + " " + (maxS + maxL));
    }
}
