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：

　　给你一个起点和一个终点，robot可以向自己所在起点的周围8个方向走，问你最短路？

题解：

　　答案就是max(abs(x2-x1),abs(y1-y2))

#include <cstdio>

#include <cstring>

#include <ctime>

#include <algorithm>

using namespace std;

typedef long long ll;

int x1,x2,y2,y1;

int main() {

    scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

    printf("%d\n",max(abs(x2-x1),abs(y1-y2)) );

    return ;

}