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 <vector>
#include <algorithm>
#include <iterator>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <cmath>
using namespace std;
#define MOD 1000000007
#define INF 1<<30
#define LINF (ll)1<<62
#define rep(i,n) for(ll i=0; i<(n); i++)
#define REP(i,a,b) for(ll i=(a); i<(b); i++)
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long ll;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef pair<ll, ll> P;
typedef vector<pair<ll, ll>> vpl;



int main(){
    ll a,b,x; cin >> a >> b >> x;
    if(a % x == 0) cout << (b/x - a/x + 1) << endl;
    else cout << (b/x - a/x) << endl;
}