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 a[2001][2001];
void solve(long long a[][2001], int n) {
  long long princ[n][n];
  long long sec[n][n];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) princ[i][j] = -1 * a[i][j];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) sec[i][j] = 0;
  long long sum;
  int row;
  int col;
  int r, c;
  for (int col = 0; col < n; col++) {
    r = 0;
    c = col;
    sum = 0;
    while (r < n && c < n) {
      sum += a[r++][c++];
    }
    r = 0;
    c = col;
    while (r < n && c < n) {
      princ[r++][c++] += sum;
    }
  }
  for (int row = 1; row < n; row++) {
    r = row;
    c = 0;
    sum = 0;
    while (r < n && c < n) {
      sum += a[r++][c++];
    }
    r = row;
    c = 0;
    while (r < n && c < n) {
      princ[r++][c++] += sum;
    }
  }
  for (int col = n - 1; col >= 0; col--) {
    r = n - 1;
    c = col;
    sum = 0;
    while (r >= 0 && c < n) {
      sum += a[r--][c++];
    }
    r = n - 1;
    c = col;
    while (r >= 0 && c < n) {
      princ[r--][c++] += sum;
    }
  }
  for (int row = n - 2; row >= 0; row--) {
    r = row;
    c = 0;
    sum = 0;
    while (r >= 0 && c < n) {
      sum += a[r--][c++];
    }
    r = row;
    c = 0;
    while (r >= 0 && c < n) {
      princ[r--][c++] += sum;
    }
  }
  long long max_odd = -1;
  long long max_even = -1;
  int x_o;
  int y_o;
  int x_e;
  int y_e;
  for (int col = 0; col < n; col++) {
    r = 0;
    c = col;
    if ((r + c) % 2) {
      while (r < n && c < n) {
        if (max_odd < princ[r][c]) {
          max_odd = princ[r][c];
          x_o = r;
          y_o = c;
        }
        r++;
        c++;
      }
    } else {
      while (r < n && c < n) {
        if (max_even < princ[r][c]) {
          max_even = princ[r][c];
          x_e = r;
          y_e = c;
        }
        r++;
        c++;
      }
    }
  }
  for (int row = 1; row < n; row++) {
    r = row;
    c = 0;
    if ((r + c) % 2) {
      while (r < n && c < n) {
        if (max_odd < princ[r][c]) {
          max_odd = princ[r][c];
          x_o = r;
          y_o = c;
        }
        r++;
        c++;
      }
    } else {
      while (r < n && c < n) {
        if (max_even < princ[r][c]) {
          max_even = princ[r][c];
          x_e = r;
          y_e = c;
        }
        r++;
        c++;
      }
    }
  }
  cout << max_odd + max_even << endl;
  cout << x_o + 1 << '\t' << y_o + 1 << '\t' << x_e + 1 << '\t' << y_e + 1
       << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int j = 0; j < n; j++)
    for (int k = 0; k < n; k++) cin >> a[j][k];
  solve(a, n);
}
