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
#include <bits/stdc++.h>
using namespace std;
int grid[2005][2005];
long long mainn[2005][2005] = {0};
long long off[2005][2005] = {0};
int n;
bool bounds(int i, int j) {
  if (i >= 1 && i <= n && j >= 1 && j <= n) return 1;
  return 0;
}
long long getMain(int i, int j) {
  if (i <= j) {
    return mainn[1][j - i + 1];
  } else {
    return mainn[i - j + 1][1];
  }
}
long long getOff(int i, int j) {
  if (i + j <= n + 1) {
    return off[1][i + j - 1];
  } else {
    return off[i + j - n][n];
  }
}
int main() {
  int i, j, t;
  cin >> n;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) scanf("%d", &grid[i][j]);
  }
  for (i = n; i >= 1; i--) {
    int x = i;
    int y = 1;
    while (bounds(x, y)) {
      mainn[i][1] += grid[x][y];
      x++;
      y++;
    }
  }
  for (i = n; i >= 1; i--) {
    int x = i;
    int y = n;
    while (bounds(x, y)) {
      off[i][n] += grid[x][y];
      x++;
      y--;
    }
  }
  for (i = n; i >= 1; i--) {
    int x = 1;
    int y = i;
    if (y != 1) {
      while (bounds(x, y)) {
        mainn[1][i] += grid[x][y];
        x++;
        y++;
      }
    }
    x = 1;
    y = i;
    if (y != n) {
      while (bounds(x, y)) {
        off[1][i] += grid[x][y];
        x++;
        y--;
      }
    }
  }
  int x1, x2, y1, y2;
  long long max1 = INT_MIN;
  long long max2 = INT_MIN;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (i & 1) {
        if (j & 1) {
          if (getOff(i, j) + getMain(i, j) - grid[i][j] > max1) {
            max1 = getOff(i, j) + getMain(i, j) - grid[i][j];
            x1 = i;
            y1 = j;
          }
        } else {
          if (getOff(i, j) + getMain(i, j) - grid[i][j] > max2) {
            max2 = getOff(i, j) + getMain(i, j) - grid[i][j];
            x2 = i;
            y2 = j;
          }
        }
      } else {
        if (j % 2 == 0) {
          if (getOff(i, j) + getMain(i, j) - grid[i][j] > max1) {
            max1 = getOff(i, j) + getMain(i, j) - grid[i][j];
            x1 = i;
            y1 = j;
          }
        } else {
          if (getOff(i, j) + getMain(i, j) - grid[i][j] > max2) {
            max2 = getOff(i, j) + getMain(i, j) - grid[i][j];
            x2 = i;
            y2 = j;
          }
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  printf("%d %d %d %d\n", x1, y1, x2, y2);
  return 0;
}
