Professor GukiZ makes a new robot. The robot are in the point with coordinates (x₁, y₁) and should go to the point (x₂, y₂). In a single step the robot can change any of its coordinates (maybe both of them) by one (decrease or increase). So the robot can move in one of the 8 directions. Find the minimal number of steps the robot should make to get the finish position.

The first line contains two integers x₁, y₁ ( - 10⁹ ≤ x₁, y₁ ≤ 10⁹) — the start position of the robot.

The second line contains two integers x₂, y₂ ( - 10⁹ ≤ x₂, y₂ ≤ 10⁹) — the finish position of the robot.

Print the only integer d — the minimal number of steps to get the finish position.

In the first example robot should increase both of its coordinates by one four times, so it will be in position (4, 4). After that robot should simply increase its y coordinate and get the finish position.

In the second example robot should simultaneously increase x coordinate and decrease y coordinate by one three times.

题意 初始位置（x1，y1） 目标位置（x2，y2）

可以行走8个方向 问最小步数

题解 max{abs（x1-x2），abs（y1-y2）}

#include<bits/stdc++.h>

using namespace std;

#define LL __int64

int main()

{

    LL a ,b, c, d;

    LL x,y;

    scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&d);

    x=abs(a-c);

    y=abs(b-d);

    if(x>y)

        printf("%I64d\n",x);

    else

        printf("%I64d\n",y);

    return 0;

}