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(raw_input())
pearls = list(map(int, raw_input().split()))
segments = []
left, right = -1, -1
while right < n - 1:
    left = right + 1
    pearl_set = set([pearls[left]])
    found = False
    for right in range(left + 1, n):
        if pearls[right] in pearl_set:
            found = True
            break
        pearl_set.add(pearls[right])
    if found:
        segments.append((left, right))
    else:
        if len(segments) == 0:
            print(-1)
            import sys
            sys.exit()
        a, b = segments.pop()
        segments.append((a, n - 1))
        break

print(len(segments))
segment_strings = map(lambda (a, b): '%s %s' % (a + 1, b + 1), segments)
print('\n'.join(segment_strings))
