Codeforce Round #217 Div2 A

A. Rook, Bishop and King

time limit per test 1 second

memory limit per test 256 megabytes

 

Little Petya is learning to play chess. He has already learned how to move a king, a rook and a bishop. Let us remind you the rules of moving chess pieces. A chessboard is 64 square fields organized into an 8 × 8 table. A field is represented by a pair of integers (r, c) — the number of the row and the number of the column (in a classical game the columns are traditionally indexed by letters). Each chess piece takes up exactly one field. To make a move is to move a chess piece, the pieces move by the following rules:

• A rook moves any number of fields horizontally or vertically.
• A bishop moves any number of fields diagonally.
• A king moves one field in any direction — horizontally, vertically or diagonally.

 

The pieces move like that

 

Petya is thinking about the following problem: what minimum number of moves is needed for each of these pieces to move from field (r₁, c₁) to field (r₂, c₂)? At that, we assume that there are no more pieces besides this one on the board. Help him solve this problem.

Input

The input contains four integers r₁, c₁, r₂, c₂ (1 ≤ r₁, c₁, r₂, c₂ ≤ 8) — the coordinates of the starting and the final field. The starting field doesn't coincide with the final one.

You can assume that the chessboard rows are numbered from top to bottom 1 through 8, and the columns are numbered from left to right 1 through 8.

Output

Print three space-separated integers: the minimum number of moves the rook, the bishop and the king (in this order) is needed to move from field (r₁, c₁) to field (r₂, c₂). If a piece cannot make such a move, print a 0 instead of the corresponding number.

Sample test(s)

Input

4 3 1 6

Output

2 1 3

Input

5 5 5 6

Output

1 0 1

#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 1000
#define INF 0x7fffffff
int n, m, k, x, y, d,a,b;
int main(){
    scanf("%d%d%d%d", &a, &b, &x, &y);
    n = abs(x - a); m = abs(y - b);
    if (n == 0 || m == 0)printf("1 "); else printf("2 ");
    if (abs(x + y - a - b) & 1)printf("0 ");
    else if (n == m)printf("1 "); else printf("2 ");
    printf("%d\n", max(n, m));
    return 0;
}