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
# http://codeforces.com/contest/616/problem/D
# Longest k-Good Segment

n, k = map(int, raw_input().split())
a = map(int, raw_input().split())

# two pointers
l = 0
r = 1

# other stuff I'm gonna use
diff_values = 1 # for a[0]
present = [0] * 1000001
present[a[0]] = 1
ans_l = 0
ans_r = 0

while r < n:
	if diff_values < k:
		if present[a[r]] == 0:
			diff_values += 1
		present[a[r]] += 1
		r += 1
	elif diff_values == k:
		if present[a[r]] != 0:
			present[a[r]] += 1
			r += 1
		else:
			if r - l > ans_r - ans_l:
				ans_l = l
				ans_r = r
			present[a[l]] -= 1
			if present[a[l]] == 0:
				diff_values -= 1
			l += 1

if r - l > ans_r - ans_l:
	ans_l = l
	ans_r = r
print ans_l + 1, ans_r