#include <bits/stdc++.h>
using namespace std;
const int N = 2e3;
int n, a[N][N];
long long L[2 * N], R[2 * N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      scanf("%d", &a[i][j]);
      L[n - i + j - 1] += a[i][j];
      R[i + j] += a[i][j];
    }
  long long res1 = -1, res2 = -1;
  int x1, x2, y1, y2;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if ((i + j) & 1) {
        long long sum = L[n - i + j - 1] + R[i + j] - a[i][j];
        if (sum > res1) {
          res1 = sum;
          x1 = i;
          y1 = j;
        }
      } else {
        long long sum = L[n - i + j - 1] + R[i + j] - a[i][j];
        if (sum > res2) {
          res2 = sum;
          x2 = i;
          y2 = j;
        }
      }
  printf("%lld\n%d %d %d %d\n", res1 + res2, x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}

### Prompt

In Cpp, your task is to solve the following problem:

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

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


Output

12
2 2 3 2

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e3;
int n, a[N][N];
long long L[2 * N], R[2 * N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) {
      scanf("%d", &a[i][j]);
      L[n - i + j - 1] += a[i][j];
      R[i + j] += a[i][j];
    }
  long long res1 = -1, res2 = -1;
  int x1, x2, y1, y2;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if ((i + j) & 1) {
        long long sum = L[n - i + j - 1] + R[i + j] - a[i][j];
        if (sum > res1) {
          res1 = sum;
          x1 = i;
          y1 = j;
        }
      } else {
        long long sum = L[n - i + j - 1] + R[i + j] - a[i][j];
        if (sum > res2) {
          res2 = sum;
          x2 = i;
          y2 = j;
        }
      }
  printf("%lld\n%d %d %d %d\n", res1 + res2, x1 + 1, y1 + 1, x2 + 1, y2 + 1);
  return 0;
}

```