The game of Chess has several pieces with curious movements. One of them is the *Queen*, which can move any number of squares in any direction: in the same line, in the same column or in any of the diagonals, as illustrated by the figure below (black dots represent positions the queen may reach in one move):

The great Chess Master Kary Gasparov invented a new type of chess problem: given the position of a queen in an empty standard chess board (that is, an 8 x 8 board) how many moves are needed so that she reaches another given square in the board?

Kary found the solution for some of those problems, but is having a difficult time to solve some others, and therefore he has asked that you write a program to solve this type of problem.

## Input

The input contains several test cases. The only line of each test case contains four integers *X _{1}*,

*Y*,

_{1}*X*and

_{2}*Y*(

_{2}*1 ≤ X*). The queen starts in the square with coordinates

_{1}, Y_{1}, X_{2}, Y_{2}≤ 8*(X*, and must finish at the square with coordinates

_{1}, Y_{1})*(X*. In the chessboard, columns are numbered from 1 to 8, from left ro right; lines are also numbered from 1 to 8, from top to bottom. The coordinates of a square in line

_{2}, Y_{2})*X*and column

*Y*are (

*X, Y*).

The end of input is indicated by a line containing four zeros, separated by spaces.

## Output

For each test case in the input your program must print a single line, containing an integer, indicating the smallest number of moves needed for the queen to reach the new position.

Sample Input | Sample Output |

4 4 6 2 3 5 3 5 5 5 4 3 0 0 0 0 |
1 0 2 |

solution here:

#include <cstdio> #include <cmath>using namespace std;int main(){ int x1, y1, x2, y2, d_x, d_y; while(scanf("%d %d %d %d", &x1, &y1, &x2, &y2) && x1 != 0 && y1 != 0 && x2 != 0 && y2 != 0){ if (x1 > x2) d_x = x1 - x2; else d_x = x2 - x1;if (y1 > y2) d_y = y1 - y2; else d_y = y2 - y1;if (x1 == x2 && y1 == y2) printf("0\n"); else if (x1 == x2 || y1 == y2 || d_x == d_y) printf("1\n"); else printf("2\n"); } return 0; }

Obrigado pela ajuda amigo!!

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Thanks for visiting!!

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