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 = input()
s = set([])
ans_i = 0
tmplist = []
alist = []

for ind, numb in enumerate(map(int, raw_input().split())):
    if not tmplist:
        tmplist.append(ind)
    if numb in s:
        tmplist.append(ind)
        s.clear()
        ans_i += 1
        alist.append(tmplist)
        tmplist = []
    else:
        s.add(numb)

if alist:
    alist[len(alist) - 1][1] = n - 1

if ans_i == 0:
    print "-1"
else:
    print ans_i
    for [a, b] in alist:
        print a + 1, b + 1
