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

n = int(raw_input())
v = map(int, sys.stdin.readline().split())

d = {}
ans = [(-1,0)]
c = 0
d[v[0]] = 0
for x in v[1:]:
	c += 1
	mx1 = 1 if x in d else -1
	mx2 = ans[c-1][0]
	mx3 = -1
	if x in d and d[x] != 0:
		mx3 = ans[d[x] -1][0] + 1

	mx = max(mx1, mx2, mx3)

	if mx3 == mx and mx > 1:
		ans.append((mx3,d[x]))
	elif mx2 == mx:
		ans.append((mx2, ans[c-1][1]))
	elif mx1 > 0:
		ans.append((1,0))
	else:
		ans.append((-1,0))

	d[x] = c

print ans[n-1][0]
if ans[n-1][0] > 0:
	s = ans[n-1][1]
	l = n
	while ans[s][0] != -1:
		print s+1, l
		l = s
		s = ans[s-1][1]

	print "1" , l


