def solution(M, A):
    l = len(A)
    distinct_slice_count = l

    # caterpillar method to solve it.
    for p in range(l - 1):
        unique_vals = set([A[p]])
        for q in range(p + 1, l):
            if A[q] in unique_vals:
                break
            else:
                unique_vals.add(A[q])
                distinct_slice_count += 1

    return distinct_slice_count