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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class cf616d {
    public static int MAX = (int) 10e6;

    public static void main(String[] args) throws IOException {
        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());
        int[] inp = new int[N]; //Input
        int[] count = new int[MAX];
        int difVal = 0;
        int l = 0;

        //Output
        int maxLen = 0;
        int maxL = 0;
        int maxR = 0;

        st = new StringTokenizer(br.readLine());
        for (int i = 0; i < N; i++) {
            int x = inp[i] = Integer.parseInt(st.nextToken());
            //Increment freq of inp[i].
            if (count[x] == 0) {
                count[x] = 1;
                difVal++;
            } else {
                count[x]++;
            }

            //If the segment doesn;t fit, trim the left side
            while (difVal > K) {
                count[inp[l]]--;
                if (count[inp[l]] == 0) {
                    difVal--;
                }
                l++;
            }

            if (i - l + 1 > maxLen) {
                maxLen = i - l + 1;
                maxL = l;
                maxR = i;
            }
        }

        System.out.println((maxL + 1) + " " + (maxR+1));


    }
}

