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;
const int M = 1e9 + 7;
int expo(int x, int y) {
  int res = 1;
  x = x % M;
  while (y > 0) {
    if (y & 1) res = (1ll * res * x) % M;
    y = y >> 1;
    x = (1ll * x * x) % M;
  }
  return res;
}
bool prime(long long int n) {
  long long int i;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, j, n;
  cin >> n;
  long long int a[n + 2][n + 2], b[n + 2][n + 2], c[n + 2][n + 2];
  memset(a, 0, sizeof(a));
  memset(b, 0, sizeof(b));
  memset(c, 0, sizeof(c));
  for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++) cin >> a[i][j];
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      b[i][j] = a[i][j] + b[i - 1][j - 1];
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      c[i][j] = a[i][j] + c[i - 1][j + 1];
    }
  }
  long long int ans = -1, ans2 = -1;
  long long int haha[4] = {1, 1, 1, 1};
  long long int right, left, right1, left1;
  long long int temp, temp2 = 0;
  for (i = 1; i <= n; i++, temp2++) {
    for (j = 1, temp = 0; j <= n; temp++, j++) {
      if (i == j) {
        left = n;
        right = n;
      } else if (j > i) {
        right = n;
        left = n - j + 1 + temp2;
      } else {
        left = n;
        right = n - i + 1 + temp;
      }
      if (i + j <= n + 1) {
        right1 = 1;
        left1 = i + j - 1;
      } else {
        left1 = n;
        right1 = i + j - n;
      }
      if ((i + j) % 2 == 0) {
        if (b[left][right] + c[left1][right1] - a[i][j] > ans) {
          ans = b[left][right] + c[left1][right1] - a[i][j];
          haha[0] = i;
          haha[1] = j;
        }
      } else if ((i + j) % 2 != 0) {
        if (b[left][right] + c[left1][right1] - a[i][j] > ans2) {
          ans2 = b[left][right] + c[left1][right1] - a[i][j];
          haha[2] = i;
          haha[3] = j;
        }
      }
    }
  }
  cout << ans + ans2 << endl;
  for (i = 0; i < 4; i++) cout << haha[i] << " ";
  return 0;
}
