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 math import ceil
from collections import defaultdict, deque

n, k = (int(i) for i in input().split())
A = [int(i) for i in input().split()]
def solve(A, n, k):
    S = set()
    d = 0
    for i in range(n):
        if A[i] not in S:
            d += 1
            S.add(A[i])
            if d == k:
                right = i
                break
    else:
        return None

    S = set()
    d = 0
    for i in reversed(range(right + 1)):
        if A[i] not in S:
            d += 1
            S.add(A[i])
            if d == k:
                left = i
                break
    return left + 1, right + 1

res = solve(A, n, k)
if res is None:
    print("-1 -1")
else:
    l, r = res
    print(f"{l} {r}")



