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
n=int(raw_input())
array = map(int,list(raw_input().split()))

d={}
ans=[]
start=1
count=0
for end,element in enumerate(array):
	if element not in d:
		d[element]=1
		# print element
	else:
		temp_list=[start,end+1]
		ans.append(temp_list)
		
		start=end+2
		count+=1
		d.clear()

if count==0:
	count=-1

print count

for index,i in enumerate(ans):
	if index==len(ans)-1:
		print i[0],max(i[1],len(array))
		continue
	print i[0],i[1]