Problem: 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

Solution (in PYTHON3):

Solution:
import sys
import math
input=sys.stdin.readline

n=int(input())
arr=list(map(int,input().split()))
ans=list()
left=-1
dc=dict()
for i in range(n):
	if(arr[i] in dc.keys() and dc[arr[i]]>left):
		ans.append([left+1,i])
		dc.pop(arr[i],None)
		left=i
	else:
		dc[arr[i]]=i
if(len(ans)==0):
	print(-1)
else:
	print(len(ans))
	for i in range(len(ans)):
		if(i==0 and i==len(ans)-1):
			print(1,n)
		elif(i==0):
			print(1,ans[i][1]+1)
		elif(i==len(ans)-1):
			print(ans[i][0]+1,n)
		else:
			print(ans[i][0]+1,ans[i][1]+1)