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, k = map(int, input().split())
a = list( map( int, sys.stdin.readline().split() ) )
ct = collections.defaultdict(int)
l, longest, numbers = 0, 0, 0
for r in range(n):
    ct[ a[r] ] += 1
    if ct[ a[r] ] == 1: numbers += 1
    while numbers > k:
        ct[ a[l] ] -= 1
        if ct[ a[l] ] == 0: numbers -= 1
        l += 1
    if r - l + 1 > longest:
        longest = r - l + 1
        ans = (l + 1, r + 1)
print('%d %d' %(ans))
