# [0..r] の剰余(mod n)の頻度
def f(r: int, n: int) -> list[int]:
    assert n > 0
    d = r // n
    res = [d] * n
    l = n * d
    for i in range(l, r+1):
        res[i % n] += 1

    return res


L, R, N = map(int, input().split())

a = f(L-1, N)
b = f(R, N)
for i in range(N):
    cnt = b[i] - a[i]
    print(cnt)