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()
arr = map(int, raw_input().strip().split())

count = 0
k = []

tmp = set()
prev = 0
for i in xrange(len(arr)):
    if arr[i] in tmp:
        tmp = set()
        count += 1
        k.append([prev+1, i + 1])
        prev = i + 1

    else:
        tmp.add(arr[i])


if count == 0:
    print -1

else:
    print count
    k[-1][-1] = len(arr)
    for i in k:
        print "{ii} {j}".format(ii = i[0], j=i[1])