## Algorithm

C. Gargari and Bishops
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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).

Input

The 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.

Output

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.

Examples
input
Copy
`41 1 1 12 1 1 01 1 1 01 0 0 1`
output
Copy
`122 2 3 2`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>

using namespace std;

int const N = 3001;
int n, g[N][N], x, y, xx, yy;
long long dp1[2 * N], dp2[2 * N], sol1 = -1, sol2 = -1;

int main() {
scanf("%d", &n);
for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j) {
scanf("%d", &g[i][j]);
dp1[i + j] += g[i][j];
dp2[i - j + n] += g[i][j];
}

for(int i = 1; i <= n; ++i)
for(int j = 1; j <= n; ++j) {
long long v = dp1[i + j] + dp2[i - j + n] - g[i][j];
if((i + j) & 1) {
if(v > sol2) {
sol2 = v;
xx = i;
yy = j;
}
} else {
if(v > sol1) {
sol1 = v;
x = i;
y = j;
}
}
}

printf("%lld\n", sol1 + sol2);
printf("%d %d %d %d\n", x, y, xx, yy);

return 0;
}``````
Copy The Code &

Input

cmd
4
1 1 1 1
2 1 1 0
1 1 1 0
1 0 0 1

Output

cmd
12
2 2 3 2