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)

#pragma region
#include <bits/stdc++.h>
using namespace std;
inline void ci(void){
  return;
}
template <typename First, typename... Rest>
void ci(First& first, Rest&... rest){
  cin >> first;
  ci(rest...);
  return;
}
inline void co(void){
  cout << "\n";
  return;
}
template <typename First, typename... Rest>
void co(First first, Rest... rest){
  cout << first << " ";
  co(rest...);
  return;
}
#define ll long long
#define ld long double
#define st string
#define dl double
#define wh while
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define tp tuple
#define ft first
#define sc second
#define ve vector
#define tos to_string
#define len length
#define rt return 0
#define br break
#define con continue
#define lcm(n, m) (n * m / gcd(n, m))
#define np next_permutation
#define repi(i, m, n) for(int i=(int)(m);i<(int)(n);i++)
#define rep(i, n) repi(i, 0, n)
#define repill(i, m, n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define repll(i, n) repill(i, 0, n)
#define revi(i, m, n) for(int i=(int)(m);i>(int)(n);i--)
#define rev(i, m) revi(i, m, 0)
#define revill(i, m, n) for(ll i=(ll)(m);i>(ll)(n);i--)
#define revll(i, n) revill(i, 0, n)
#define sz(v) ((int)(v).size()) 
#define each(i, n) for(auto&& i: n)
#define all(n) (n.begin(), n.end())
#define allr(n) (n.rbegin(), n.rend())
#define sp(n) cout << fixed << setprecision(n)
#pragma endregion
// sqrt ルート, int 2*10^9, gcd　最大公約数
// __builtin_popcount bitの個数

int main() {
  ll a, b, x;
  ci(a, b, x);
  if(a == 0){
    co(b / x + 1);
  }
  else{
    co(b / x - (a - 1) / x);
  }
}