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
def find_longest_segment(arr, k):
    
    freq = {}
    freq[arr[0]] = 1
    lower_bound, upper_bound = 1, 1
    j = 0
    max_length = 1
    unique = 1
    
    for i in xrange(1, len(arr)):
        if arr[i] not in freq:
            while unique == k and j < i:
                freq[arr[j]] -= 1
                if freq[arr[j]] == 0:
                    del freq[arr[j]]
                    unique -= 1
                j += 1
            freq[arr[i]] = 0
            unique += 1
        freq[arr[i]] += 1
        
        if max_length <= i - j + 1:
            max_length = i - j + 1
            lower_bound, upper_bound = j + 1, i + 1
            
    return "{} {}".format(lower_bound, upper_bound)

n, k = map(int, raw_input().strip().split(' '))
arr = map(int, raw_input().strip().split(' '))

print find_longest_segment(arr, k)
# your code goes here