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>).
n,k = map(int,input().split())
a = list(map(int,input().split()))

count = 0
b = {}
for i in range(n):
    if a[i] in b:
        b[a[i]] += 1
    else:
        b[a[i]] =1
    if b[a[i]] == 1:
        count += 1
    if count == k:
        break
#print("i=", i)
#print(b)


for j in range(n):
  if a[j] in b:
    b[a[j]] -= 1
#    print("j=", j)
#    print(b)
  if b[a[j]] == 0:
    break 
#  print("j=",j)
        
if count != k:
    print("-1 -1")
else:
  if n == 1:
    print(1,1)
  elif n == 2 and count == 2:
    print(1,2)
  else:
    print(j+1,i+1)  