Problem:
Score : 200 points

Problem Statement

AtCoDeer the deer found two rectangles lying on the table, each with height 1 and width W.
If we consider the surface of the desk as a two-dimensional plane, the first rectangle covers the vertical range of [0,1] and the horizontal range of [a,a+W], and the second rectangle covers the vertical range of [1,2] and the horizontal range of [b,b+W], as shown in the following figure:

AtCoDeer will move the second rectangle horizontally so that it connects with the first rectangle.
Find the minimum distance it needs to be moved.

Constraints

All input values are integers.

1≤W≤10^5

1≤a,b≤10^5

Input

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

W a b

Output

Print the minimum distance the second rectangle needs to be moved.

Sample Input 1

3 2 6

Sample Output 1

1

This input corresponds to the figure in the statement. In this case, the second rectangle should be moved to the left by a distance of 1.

Sample Input 2

3 1 3

Sample Output 2

0

The rectangles are already connected, and thus no move is needed.

Sample Input 3

5 10 1

Sample Output 3

4

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <numeric>
#include <iostream>
#include <string>
#include <cmath>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
typedef long long ll;
using vi = vector<int>; // intの1次元の型に vi という別名をつける
using vvi = vector<vi>; // intの2次元の型に vvi という別名をつける
using vl = vector<ll>;  // long longの1次元の型に vl という別名をつける
using vvl = vector<vl>; // long longの2次元の型に vvl という別名をつける
int MOD = 1000000007; //10^9 + 7

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

// 無限大の値
const ll  lINF = 1LL << 60;
const int iINF = 1 << 29;

int main(void){
  int w,a,b;
  cin >> w >> a >> b;

  if((a<=b && b<=a+w) || (a<=b+w && b+w<=a+w)) cout << 0 << endl;
  else if(abs(b-(a+w))<=abs(a-(b+w))) cout << abs(b-(a+w)) << endl;
  else cout << abs(a-(b+w)) << endl;
  return 0;
}