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 collections import Counter
n, k=[int(v) for v in input().split()]
w=[int(v) for v in input().split()]
q=Counter(w)
c=len(list(q.keys()))
a, b=0, n-1
if c<k:
    print(-1, -1)
else:
    while c>=k:
        if q[w[a]]==1:
            if c==k:
                pass
            else:
                q[w[a]]-=1
                c-=1
                a+=1
                continue
        else:
            q[w[a]]-=1
            a+=1
            continue
        if q[w[b]]==1:
            if c==k:
                print(a+1, b+1)
                break
            else:
                q[w[b]]-=1
                c-=1
                b-=1
        else:
            q[w[b]]-=1
            b-=1