#include <bits/stdc++.h>
using namespace std;
int num[2005][2005];
long long sumN[2005 * 2], sumM[2005 * 2];
int n;
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(sumN, 0, sizeof(sumN));
    memset(sumM, 0, sizeof(sumM));
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) {
        scanf("%d", &num[i][j]);
        sumN[i + j] += num[i][j];
        sumM[i - j + n] += num[i][j];
      }
    long long maxOdd = -1, maxEvent = -1, s;
    int x1, x2, y1, y2;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) {
        if ((i + j) & 1) {
          if (maxOdd < (s = sumN[i + j] + sumM[i - j + n] - num[i][j])) {
            maxOdd = s;
            x1 = i;
            y1 = j;
          }
        } else {
          if (maxEvent < (s = sumN[i + j] + sumM[i - j + n] - num[i][j])) {
            maxEvent = s;
            x2 = i;
            y2 = j;
          }
        }
      }
    printf("%lld\n", maxOdd + maxEvent);
    printf("%d %d %d %d\n", x1, y1, x2, y2);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int num[2005][2005];
long long sumN[2005 * 2], sumM[2005 * 2];
int n;
int main() {
  while (scanf("%d", &n) != EOF) {
    memset(sumN, 0, sizeof(sumN));
    memset(sumM, 0, sizeof(sumM));
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) {
        scanf("%d", &num[i][j]);
        sumN[i + j] += num[i][j];
        sumM[i - j + n] += num[i][j];
      }
    long long maxOdd = -1, maxEvent = -1, s;
    int x1, x2, y1, y2;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= n; ++j) {
        if ((i + j) & 1) {
          if (maxOdd < (s = sumN[i + j] + sumM[i - j + n] - num[i][j])) {
            maxOdd = s;
            x1 = i;
            y1 = j;
          }
        } else {
          if (maxEvent < (s = sumN[i + j] + sumM[i - j + n] - num[i][j])) {
            maxEvent = s;
            x2 = i;
            y2 = j;
          }
        }
      }
    printf("%lld\n", maxOdd + maxEvent);
    printf("%d %d %d %d\n", x1, y1, x2, y2);
  }
  return 0;
}

```