You've got an array *a*, consisting of *n* integers: *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub>. Your task is to find a minimal by inclusion segment [*l*, *r*] (1 ≤ *l* ≤ *r* ≤ *n*) such, that among numbers *a*<sub class="lower-index">*l*</sub>,  *a*<sub class="lower-index">*l* + 1</sub>,  ...,  *a*<sub class="lower-index">*r*</sub> there are exactly *k* distinct numbers.

Segment [*l*, *r*] (1 ≤ *l* ≤ *r* ≤ *n*; *l*, *r* are integers) of length *m* = *r* - *l* + 1, satisfying the given property, is called minimal by inclusion, if there is no segment [*x*, *y*] satisfying the property and less then *m* in length, such that 1 ≤ *l* ≤ *x* ≤ *y* ≤ *r* ≤ *n*. Note that the segment [*l*, *r*] doesn't have to be minimal in length among all segments, satisfying the given property.
The first line contains two space-separated integers: *n* and *k* (1 ≤ *n*, *k* ≤ 10<sup class="upper-index">5</sup>). The second line contains *n* space-separated integers *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> — elements of the array *a* (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>).
from sys import stdin,stdout
stdin.readline
def mp(): return list(map(int, stdin.readline().strip().split()))
def it():return int(stdin.readline().strip())
from collections import defaultdict as dd,Counter as C,deque
from math import ceil,gcd,sqrt,factorial,log2,floor	
# from bisect import bisect_right as br,bisect_left as bl
import heapq

n,k = mp()
l = mp()
if len(set(l)) < k:
	print(-1,-1)
	exit()
if n == 1 or len(set(l))==1:
	print(1,1)
	exit()
d = {}
c = 1
for i in l:
	if len(d) == k:
		break
	d[i] = c
	c += 1
if len(d) == k:
	print(min(d.values()) , max(d.values()))
else:
	print(-1,-1)

