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 = list(map(int, input().split()))
a = list(map(int, input().split()))
l = 1
r = n
old_e = a[0]
count = 1
no_duplicate = 0
distinc_a = [old_e]
for index, e in enumerate(a, start=1):
    if index == 1:
        continue
    if e == old_e:
        if count == 1:
            l = index
    else:
        old_e = e
        for distinc_e in distinc_a:
            if distinc_e != e:
                pass
            else:
                break
        else:
            distinc_a.append(e)
            count += 1
        
    if count == k:
        r = index
        break
else:
    if count < k:
        l = r = -1
    
print(l, r)
    