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.InputStreamReader;
import java.util.HashSet;

public class Edu6B {
	public static void main(String[] args) throws Exception {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bf.readLine());
		String l[] = bf.readLine().split(" ");
		int nums[] = new int[n];
		for (int i = 0; i < nums.length; i++) {
			nums[i] = Integer.parseInt(l[i]);
		}
		HashSet<Integer> lastInd = new HashSet<>();
		StringBuffer sb = new StringBuffer();
		int ans = 0, last = 0;
		int segs[] = new int[n + 1];
		int ind = 1;
		for (int i = 0; i < nums.length; i++) {
			if (lastInd.contains(nums[i])) {
				ans++;
				segs[last] = i;
				last = i + 1;
				lastInd.clear();
			} else {
				lastInd.add(nums[i]);
			}
		}
		if (ans == 0) {
			System.out.println(-1);
		} else {
			int i = 0;
			sb.append(ans + "\n");
			while (segs[i] != 0) {
				if (segs[segs[i] + 1] == 0) {
					sb.append((i + 1) + " " + (n) + "\n");
					break;
				} else {
					sb.append((i + 1) + " " + (segs[i] + 1) + "\n");
					i = segs[i] + 1;
				}
			}
			System.out.print(sb);
		}
	}
}
