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
__author__ = 'aste'


def main():
    n = int(raw_input())
    a = map(int, raw_input().split())
    res = []
    s = set()
    s.add(a[0])
    i = 0
    j = 1
    while j < n:
        if a[j] in s:
            res.append((i, j))
            i = j + 1
            s = set()
        else:
            s.add(a[j])
        j += 1
    if res:
        if i < n:
            si, sj = res.pop()
            res.append((si, n - 1))
    if res:
        print len(res)
        print "\n".join(["%d %d" % (k + 1, v + 1) for k, v in res])
    else:
        print -1

main()