Gargari is jealous that his friend Caisa won the game from the previous problem. He wants to prove that he is a genius.
He has a n?×?n chessboard. Each cell of the chessboard has a number written on it. Gargari wants to place two bishops on the chessboard in such a way that there is no cell that is attacked by both of them. Consider a cell with number x written on it, if this cell is attacked by one of the bishops Gargari will get x dollars for it. Tell Gargari, how to place bishops on the chessboard to get maximum amount of money.
We assume a cell is attacked by a bishop, if the cell is located on the same diagonal with the bishop (the cell, where the bishop is, also considered attacked by it).
InputThe first line contains a single integer n (2?≤?n?≤?2000). Each of the next n lines contains n integers aij (0?≤?aij?≤?109) ― description of the chessboard.
On the first line print the maximal number of dollars Gargari will get. On the next line print four integers: x1,?y1,?x2,?y2 (1?≤?x1,?y1,?x2,?y2?≤?n), where xi is the number of the row where the i-th bishop should be placed, yi is the number of the column where the i-th bishop should be placed. Consider rows are numbered from 1 to n from top to bottom, and columns are numbered from 1 to n from left to right.
If there are several optimal solutions, you can print any of them.
Sample test(s) input4 1 1 1 1 2 1 1 0 1 1 1 0 1 0 0 1output
12 2 2 3 2
题意：在n*n的矩阵中选两个点，要求这两个点所覆盖的点的总和最大，且它们分别覆盖的点不能有相同的，一给点能被另一个点覆盖当且仅当它们在同一斜线上
思路：既然要求覆盖的点不能有相同的，假设一个点为（i ， j ），令v=i+j，那么这两个点的v值一定一个是奇数，一个是偶数，否则无论怎么选都会有相同的点被覆盖，所以只
需要预处理出矩阵每个斜行数（左斜，右斜）的总和，然后遍历一遍矩阵的点，分点的v值为奇数和偶数分别记录最大值即可，很容易想到这个思路，但是处理起来觉得 略麻烦，看来还是我的代码能力不够，sad~