#include <bits/stdc++.h>
using namespace std;
long long mp[2010][2010];
long long sum1[2010][2010], sum2[2010][2010];
int n;
int check(int x, int y, int xx, int yy) {
  int remy = abs(yy - y);
  int remx = abs(xx - x);
  if (remy == remx) return 0;
  return 1;
}
long long getsum(int i, int j) {
  long long s1 = 0, s2 = 0;
  int x1 = j - 1 + i, y1 = 1;
  if (x1 <= n) s1 = sum2[x1][y1];
  int x2 = n, y2 = j - n + i;
  if (y2 >= 1) s1 = sum2[x2][y2];
  x1 = n - j + i, y1 = n;
  if (x1 <= n) s2 = sum1[x1][y1];
  x2 = n, y2 = j + n - i;
  if (y2 <= n) s2 = sum1[x2][y2];
  return s1 + s2 - mp[i][j];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) scanf("%I64d", &mp[i][j]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) sum1[i][j] = sum1[i - 1][j - 1] + mp[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = n; j >= 1; j--) sum2[i][j] = sum2[i - 1][j + 1] + mp[i][j];
  long long maxv1 = sum2[1][1] + sum1[n][n] - 1;
  long long maxv2 = sum2[2][1] + sum1[n - 1][n] - 1;
  int x = 1, y = 1, xx = 1, yy = 2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long sum = getsum(i, j);
      if ((i + j) % 2 == 1 && sum > maxv2) {
        x = i;
        y = j;
        maxv2 = sum;
      }
      if ((i + j) % 2 == 0 && sum > maxv1) {
        maxv1 = sum;
        xx = i;
        yy = j;
      }
    }
  }
  printf("%I64d\n%d %d %d %d\n", maxv1 + maxv2, x, y, xx, yy);
}

### Prompt

Please provide a CPP coded solution 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;
long long mp[2010][2010];
long long sum1[2010][2010], sum2[2010][2010];
int n;
int check(int x, int y, int xx, int yy) {
  int remy = abs(yy - y);
  int remx = abs(xx - x);
  if (remy == remx) return 0;
  return 1;
}
long long getsum(int i, int j) {
  long long s1 = 0, s2 = 0;
  int x1 = j - 1 + i, y1 = 1;
  if (x1 <= n) s1 = sum2[x1][y1];
  int x2 = n, y2 = j - n + i;
  if (y2 >= 1) s1 = sum2[x2][y2];
  x1 = n - j + i, y1 = n;
  if (x1 <= n) s2 = sum1[x1][y1];
  x2 = n, y2 = j + n - i;
  if (y2 <= n) s2 = sum1[x2][y2];
  return s1 + s2 - mp[i][j];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) scanf("%I64d", &mp[i][j]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) sum1[i][j] = sum1[i - 1][j - 1] + mp[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = n; j >= 1; j--) sum2[i][j] = sum2[i - 1][j + 1] + mp[i][j];
  long long maxv1 = sum2[1][1] + sum1[n][n] - 1;
  long long maxv2 = sum2[2][1] + sum1[n - 1][n] - 1;
  int x = 1, y = 1, xx = 1, yy = 2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long sum = getsum(i, j);
      if ((i + j) % 2 == 1 && sum > maxv2) {
        x = i;
        y = j;
        maxv2 = sum;
      }
      if ((i + j) % 2 == 0 && sum > maxv1) {
        maxv1 = sum;
        xx = i;
        yy = j;
      }
    }
  }
  printf("%I64d\n%d %d %d %d\n", maxv1 + maxv2, x, y, xx, yy);
}

```