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
s1 = input().split()
length = int(s1[0])
max_k = int(s1[1])
s2 = input()

arr = s2.split()
start = 0
end = 0
dist = dict()
s = start
max_l = 0
for i in range(length):
    cur = arr[i]
    if cur not in dist:
        dist[cur] = 1
    else:
        dist[cur] += 1
    k = len(dist)
    l = i - s
    if k > max_k:
        to_remove = arr[s]
        if dist[to_remove] == 1:
            dist.pop(to_remove)
        else:
            dist[to_remove] -= 1
        l -= 1
        s += 1
    if l > max_l:
        start = s
        max_l = l
        end = i


print(start + 1, end + 1)
