# [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


# 区間 [0..r] で n で割った余りが m となるものの個数を返す
def g(r: int, n: int, m: int) -> int:
    assert n > 0
    assert 0 <= m < n
    res = r // n
    if m <= r % n:
        res += 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 = g(R, N, i) - g(L-1, N, i)
    print(cnt)