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
def solve(n, k, a):
	l, r, cur, answer_l, answer_r = 0, 0, 1, 0, 0

	cnt = [0 for _ in range(1000010)]
	cnt[a[0]] = 1

	while r + 1 < n:
		r += 1
		cnt[a[r]] += 1

		if cnt[a[r]] == 1:
			cur += 1

		if cur > k:
			if r - l - 1 > answer_r - answer_l:
				answer_l, answer_r = l, r - 1

		while cur > k:
			cnt[a[l]] -= 1
			
			if cnt[a[l]] == 0:
				cur -= 1

			l += 1

	if r - l > answer_r - answer_l:
		answer_l, answer_r = l, r

	print(answer_l + 1, answer_r + 1)


if __name__ == '__main__':
	n, k = map(int, input().split())

	a = list(map(int, input().split()))

	solve(n, k, a)
