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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <cmath>
//ranker
using namespace std;

#define REPS(i, a, n) for (int i = (a); i < (n); ++i)
#define REP(i, n) REPS(i, 0, n)
#define RREP(i, n) REPS(i, 1, n + 1)
#define DEPS(i, a, n) for (int i = (a); i >= n; --i)
#define DEP(i, n) DEPS(i, n, 0)

using ll = long long;
using ull = unsigned long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using pis = pair<int, string>;
using psi = pair<string, int>;
using D = double;


int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);

  long long int a, b, a1, b1, x;

  cin >> a >> b >> x;

  b1 = b/x;
  a1 = a/x;

  if(a != 0 && a%x != 0) cout << b1-a1 << endl;
  if(a != 0 && a%x == 0) cout << b1-a1+1 << endl;
  if(a == 0) cout << b1+1 << endl;

  return 0;
}