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
n = int(input())
index = 1
memory = set()
counter = 0
segments = [[0, 0]]
for number in input().split():
    if number not in memory:
        memory.add(number)
    else:
        counter += 1
        segments.append([segments[-1][1] + 1, index])
        memory = set()
    index += 1
segments[-1][1] = n
if counter != 0:
    print(counter)
    segments.remove([0, 0])
    print('\n'.join('{0} {1}'.format(p, q) for (p, q) in segments))
else:
    print(-1)