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 arr[2007][2007], sum1[2007][2007], sum2[2007][2007];
int main() {
  long long n, i, j, k, l, x, y;
  cin >> n;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      scanf("%lld", &x);
      arr[i][j] = x;
    }
  }
  long long s1 = 0, s2 = 0;
  for (i = 1; i <= n; i++) {
    s1 = 0;
    for (j = 1; j <= i; j++) {
      s1 += arr[j][i - j + 1];
    }
    for (j = 1; j <= i; j++) {
      sum1[j][i - j + 1] = s1;
    }
  }
  for (i = n; i >= 1; i--) {
    s1 = 0;
    for (j = n; j >= i; j--) {
      s1 += arr[j][i - j + n];
    }
    for (j = n; j >= i; j--) {
      sum1[j][i - j + n] = s1;
    }
  }
  for (i = 1; i <= n; i++) {
    s1 = 0;
    for (j = 1; j <= i; j++) {
      s1 += arr[j][n - i + j];
    }
    for (j = 1; j <= i; j++) {
      sum2[j][n - i + j] = s1;
    }
  }
  for (i = n; i >= 1; i--) {
    s1 = 0;
    for (j = n; j >= i; j--) {
      s1 += arr[j][j - i + 1];
    }
    for (j = n; j >= i; j--) {
      sum2[j][j - i + 1] = s1;
    }
  }
  long long max1 = 0, max2 = 0;
  long long sum;
  long long idx1 = 1, idx2 = 1, idy1 = 1, idy2 = 2;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2LL) {
        sum = sum1[i][j] + sum2[i][j] - arr[i][j];
        if (sum > max1) {
          max1 = sum;
          idx1 = i;
          idy1 = j;
        }
      } else {
        sum = sum1[i][j] + sum2[i][j] - arr[i][j];
        if (sum > max2) {
          max2 = sum;
          idx2 = i;
          idy2 = j;
        }
      }
    }
  }
  cout << max1 + max2 << endl;
  cout << idx1 << " " << idy1 << " " << idx2 << " " << idy2 << endl;
  return 0;
}
