Problem: 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

Solution (in PYTHON):

Solution:
def indexs(nums, k, numsCnt):
	cnt, begin, i = 0, 0, 0
	longest, beginAns, endAns = k, 0, k
	d = {}

	while(i < numsCnt):
		while(i < numsCnt and cnt <= k):
			if((not nums[i] in d) or d[nums[i]] == 0):
				d[nums[i]] = 1
				cnt += 1
			else:
				d[nums[i]] += 1
			i += 1

		# print("i = %d, begin = %d, longest = %d"%(i, begin, longest))
		if(i - 1 - begin > longest):
			beginAns = begin
			endAns = i - 1
			longest = i - 1 - begin
		
		while(begin < numsCnt and cnt > k):
			d[nums[begin]] -= 1
			if(d[nums[begin]] == 0):
				cnt -= 1
			begin += 1 

	if(i - begin > longest):
		beginAns = begin
		endAns = i

	return (beginAns, endAns)

if(__name__ == "__main__"):
	numsCnt, k = [int(tmp) for tmp in raw_input().split(" ")]
	nums = [int(tmp) for tmp in raw_input().split(" ")]
	begin, end = indexs(nums, k, numsCnt);
	print("%d %d"%(begin + 1, end))