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() )
A = map(int, raw_input().split())

M = {}
left = 1
IP = []
# traverse A
SZ = len(A)
for i in range(1, SZ + 1):
	if A[i-1] in M:
		M.clear()
		IP.append([left, i])
		left = i + 1
	else:
		M[ A[i-1] ] = 1

if IP:
	L = len(IP)
	# R for last seg
	IP[-1][1] = SZ
	print L
	for i in range(L):
		print IP[i][0], IP[i][1]
else:
	print -1
