Problem:
Score : 200 points

Problem Statement

You are given nonnegative integers a and b (a ≤ b), and a positive integer x.
Among the integers between a and b, inclusive, how many are divisible by x?

Constraints

0 ≤ a ≤ b ≤ 10^{18}

1 ≤ x ≤ 10^{18}

Input

The input is given from Standard Input in the following format:

a b x

Output

Print the number of the integers between a and b, inclusive, that are divisible by x.

Sample Input 1

4 8 2

Sample Output 1

3

There are three integers between 4 and 8, inclusive, that are divisible by 2: 4, 6 and 8.

Sample Input 2

0 5 1

Sample Output 2

6

There are six integers between 0 and 5, inclusive, that are divisible by 1: 0, 1, 2, 3, 4 and 5.

Sample Input 3

9 9 2

Sample Output 3

0

There are no integer between 9 and 9, inclusive, that is divisible by 2.

Sample Input 4

1 1000000000000000000 3

Sample Output 4

333333333333333333

Watch out for integer overflows.

Solution in Python (3.4.3) (judge verdict: Accepted)

import sys
import math
import collections
import itertools
import array
import inspect

# Set max recursion limit
sys.setrecursionlimit(1000000)


# Debug output
def chkprint(*args):
    names = {
        id(v): k
        for k, v in inspect.currentframe().f_back.f_locals.items()
    }
    print(', '.join(
        names.get(id(arg), '???') + ' = ' + repr(arg) for arg in args))


# Binary converter
def to_bin(x):
    return bin(x)[2:]


def li_input():
    return [int(_) for _ in input().split()]


def gcd(n, m):
    if n % m == 0:
        return m
    else:
        return gcd(m, n % m)


def gcd_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = gcd(v, L[i])

    return v


def lcm(n, m):
    return (n * m) // gcd(n, m)


def lcm_list(L):
    v = L[0]

    for i in range(1, len(L)):
        v = lcm(v, L[i])

    return v


# Width First Search (+ Distance)
def wfs_d(D, N, K):
    """
    D: 隣接行列(距離付き)
    N: ノード数
    K: 始点ノード
    """

    dfk = [-1] * (N + 1)
    dfk[K] = 0

    cps = [(K, 0)]
    r = [False] * (N + 1)
    r[K] = True
    while len(cps) != 0:
        n_cps = []
        for cp, cd in cps:
            for i, dfcp in enumerate(D[cp]):
                if dfcp != -1 and not r[i]:
                    dfk[i] = cd + dfcp
                    n_cps.append((i, cd + dfcp))
                    r[i] = True

        cps = n_cps[:]

    return dfk


# Depth First Search (+Distance)
def dfs_d(v, pre, dist):
    """
    v:  現在のノード
    pre: １つ前のノード
    dist: 現在の距離

    以下は別途用意する
    D: 隣接リスト(行列ではない)
    D_dfs_d: dfs_d関数で用いる，始点ノードから見た距離リスト
    """

    global D
    global D_dfs_d

    D_dfs_d[v] = dist

    for next_v, d in D[v]:
        if next_v != pre:
            dfs_d(next_v, v, dist + d)

    return


def sigma(N):
    ans = 0
    for i in range(1, N + 1):
        ans += i
    return ans


class Combination:
    def __init__(self, n, mod):

        g1 = [1, 1]
        g2 = [1, 1]
        inverse = [0, 1]
        for i in range(2, n + 1):
            g1.append((g1[-1] * i) % mod)
            inverse.append((-inverse[mod % i] * (mod // i)) % mod)
            g2.append((g2[-1] * inverse[-1]) % mod)
        self.MOD = mod
        self.N = n
        self.g1 = g1
        self.g2 = g2
        self.inverse = inverse

    def __call__(self, n, r):
        if (r < 0 or r > n):
            return 0
        r = min(r, n - r)
        return self.g1[n] * self.g2[r] * self.g2[n - r] % self.MOD


# --------------------------------------------

dp = None


def f(n, x):
    if n == -1:
        return 0

    return n // x + 1


def main():
    a, b, x = li_input()

    print(f(b, x) - f(a - 1, x))


main()