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())
w = map(int, raw_input().split())
p = set(w)
if(len(w)==len(p)):
	print -1
	exit(0)
ans = []
temp = set([])
lol = []
k = 0
for i in range(n):
	temp.add(w[i])
	if(lol==[]):
		lol.append(i+1)
	if(k==len(temp)):
		lol.append(i+1)
		ans.append(lol)
		temp = set([])
		lol = []
		k = 0
	else:
		k+=1

if(k!=0):
	ans[-1][1] = n

print len(ans)
for i in ans:
	for j in i:
		print j,
	print