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
import sys
import collections
n, a = int( input() ), list( map( int, sys.stdin.readline().split() ) )
f = collections.defaultdict(int)
ans = list()
prev, curr = 1, 1
for i in range(n):
    f[ a[i] ] += 1
    if f[ a[i] ] == 2:
        ans.append( (prev, curr) )
        for j in range(prev - 1, curr):
            f[ a[j] ] -= 1
        prev = curr + 1
    curr += 1
if len(ans) == 0:
    print('-1')
else:
    l = ans.pop()
    ans.append( (l[0], n) )
    print(len(ans))
    for x, y in ans:
        sys.stdout.write( str(x) + ' ' + str(y) + '\n' )
