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 add(elements, to_add):
    if to_add not in elements:
        elements[to_add] = 0
    elements[to_add] += 1
 
 
def remove(elements, to_remove):
    elements[to_remove] -= 1
    if elements[to_remove] == 0:
        del elements[to_remove]
 
 
_, limit = [int(i) for i in input().split()]
arr = [int(i) for i in input().split()]
 
longest_good = 0
good_seg = [-1, -1]
 
left = 0
elements = {}
for right in range(len(arr)):
    add(elements, arr[right])
    while len(elements) > limit:
        remove(elements, arr[left])
        left += 1
    if right - left + 1 > longest_good:
        longest_good = right - left + 1
        good_seg = [left + 1, right + 1]
print(good_seg[0], good_seg[1])
