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;
long long s1[2005][2005], s2[2005][2005], a[2005][2005];
int main() {
  int n, x, y, u, v;
  long long maxVal, smaxVal;
  scanf("%d", &n);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) scanf("%I64d", &a[i][j]);
  for (int i = 0; i < n; i++) {
    int j = i, k = 0;
    long long s = 0;
    while (j < n) {
      s += a[j++][k++];
    }
    j = i, k = 0;
    while (j < n) {
      s1[j++][k++] = s;
    }
  }
  for (int i = 1; i < n; i++) {
    int j = i, k = 0;
    long long s = 0;
    while (j < n) {
      s += a[k++][j++];
    }
    j = i, k = 0;
    while (j < n) {
      s1[k++][j++] = s;
    }
  }
  for (int i = 0; i < n; i++) {
    int j = i, k = 0;
    long long s = 0;
    while (j >= 0) {
      s += a[k++][j--];
    }
    j = i, k = 0;
    while (j >= 0) s2[k++][j--] = s;
  }
  for (int i = 1; i < n; i++) {
    int j = i, k = n - 1;
    long long s = 0;
    while (j < n) {
      s += a[j++][k--];
    }
    j = i, k = n - 1;
    while (j < n) s2[j++][k--] = s;
  }
  maxVal = smaxVal = -1000000000;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (s1[i][j] + s2[i][j] - a[i][j] > maxVal) {
          maxVal = s1[i][j] + s2[i][j] - a[i][j];
          x = i;
          y = j;
        }
      } else {
        if (s1[i][j] + s2[i][j] - a[i][j] > smaxVal) {
          smaxVal = s1[i][j] + s2[i][j] - a[i][j];
          u = i;
          v = j;
        }
      }
    }
  }
  printf("%I64d\n", maxVal + smaxVal);
  printf("%d %d %d %d\n", x + 1, y + 1, u + 1, v + 1);
  return 0;
}
