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 mini(int a, int b) { return a < b ? a : b; }
long long lmini(long long a, long long b) { return a < b ? a : b; }
int maxi(int a, int b) { return a > b ? a : b; }
long long lmaxi(long long a, long long b) { return a > b ? a : b; }
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lgcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int lcm(int a, int b) { return a * (b / gcd(a, b)); }
long long llcm(long long a, long long b) { return a * (b / gcd(a, b)); }
long long mat[2001][2001], n, m1 = -1, m2 = -1;
vector<long long> v1, v2;
int main() {
  ios_base::sync_with_stdio(false);
  cin.clear();
  cin >> n;
  v1.resize(2 * n, 0);
  v2.resize(2 * n, 0);
  long long i, j;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      cin >> mat[i][j];
    }
  }
  long long c = 1;
  i = 1;
  j = 1;
  while (i <= n) {
    long long sum = 0;
    long long k = j, m = i;
    while (m > 0) {
      sum += mat[m][k];
      m--;
      k++;
    }
    v1[i] = sum;
    i++;
    j = 1;
  }
  j = 2;
  c = i;
  i = n;
  while (j <= n) {
    long long sum = 0;
    long long k = j, m = i;
    while (k <= n) {
      sum += mat[m][k];
      k++;
      m--;
    }
    v1[c] = sum;
    c++;
    i = n;
    j++;
  }
  c = 1;
  i = n;
  j = 1;
  while (i > 0) {
    long long sum = 0;
    long long k = j, m = i;
    while (m <= n) {
      sum += mat[m][k];
      m++;
      k++;
    }
    v2[c] = sum;
    c++;
    i--;
    j = 1;
  }
  j = n;
  i = n - 1;
  while (i > 0) {
    long long sum = 0;
    long long k = j, m = i;
    while (m > 0) {
      sum += mat[m][k];
      k--;
      m--;
    }
    v2[c] = sum;
    c++;
    i--;
    j = n;
  }
  long long f1 = 1, f2 = 1, a, b, c1, d;
  for (i = 1; i <= n; i++) {
    if (f1 == 0)
      f1 = 1;
    else
      f1 = 0;
    if (n % 2 == 1) f2 = !f2;
    for (j = 1; j <= n; j++) {
      if (f2 == 1)
        f2 = 0;
      else
        f2 = 1;
      if (f1 == 0) {
        if (f2 == 0) {
          if (v1[i + j - 1] + v2[n + j - i] - mat[i][j] > m1) {
            m1 = v1[i + j - 1] + v2[n + j - i] - mat[i][j];
            a = i;
            b = j;
          }
        } else {
          if (v1[i + j - 1] + v2[n + j - i] - mat[i][j] > m2) {
            m2 = v1[i + j - 1] + v2[n + j - i] - mat[i][j];
            c1 = i;
            d = j;
          }
        }
      } else {
        if (f2 == 0) {
          if (v1[i + j - 1] + v2[n + j - i] - mat[i][j] > m2) {
            m2 = v1[i + j - 1] + v2[n + j - i] - mat[i][j];
            c1 = i;
            d = j;
          }
        } else {
          if (v1[i + j - 1] + v2[n + j - i] - mat[i][j] > m1) {
            m1 = v1[i + j - 1] + v2[n + j - i] - mat[i][j];
            a = i;
            b = j;
          }
        }
      }
    }
  }
  cout << m1 + m2 << endl;
  cout << a << " " << b << " " << c1 << " " << d << endl;
  return 0;
}
