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>).
maxa = 10**5 + 1
iarr = list(map(int,input().split()))
n = iarr[0]
k = iarr[1]
arr = list(map(int,input().split()))
freq = [0 for i in range(maxa)]
for i in range(n):
	freq[arr[i]]+=1
dis = 0
for i in range(maxa):
	if freq[i]>0:
		dis+=1
#print(dis)
if dis<k:
	print(-1,-1)
else:
	i=0
	while True:
		#print(i,dis)
		freq[arr[i]]-=1
		if freq[arr[i]]==0:
			dis-=1
		if dis==k-1:
			break
		i+=1
	freq[i]+=1
	l = i

	i=n-1
	while True:
		freq[arr[i]]=-1
		if freq[arr[i]]==0:
			dis-=1
		if dis==k-1:
			break
		i-=1
	freq[i]+=1
	r = i
	print(l+1,r+1)










