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
#! /usr/bin/python

n = int(raw_input())
row = [int(x) for x in raw_input().split()]
result = []
l = 0
r = 0
currSet = set(row[l:r])
while r<n:
    if row[r] in currSet:
        result.append([l+1, r+1])
        l = r+1
        r = r+1
        currSet.clear()
    else:
        currSet.add(row[r])
        r = r+1
if len(result) == 0:
    print -1
else:
    print len(result)
    result[-1][1] = n
    for r in result:
        print " ".join(str(i) for i in r)
