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 N = 55;
const long long M = 1e9 + 7;
const long long inf = 1e18;
int m;
bool check(int n) {
  if (n >= 0 and n < m) return 1;
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  m = n;
  long long a[n][n];
  long long diag[n][n];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) scanf("%lld", &a[i][j]);
  }
  for (int i = 0; i < n; i++) {
    int row = i, col = 0;
    long long val = 0;
    while (check(row) and check(col)) {
      val += a[row][col];
      row++;
      col++;
    }
    row = i, col = 0;
    while (check(row) and check(col)) {
      diag[row][col] = val;
      row++;
      col++;
    }
  }
  for (int i = 1; i < n; i++) {
    int row = 0, col = i;
    long long val = 0;
    while (check(row) and check(col)) {
      val += a[row][col];
      row++;
      col++;
    }
    row = 0, col = i;
    while (check(row) and check(col)) {
      diag[row][col] = val;
      row++;
      col++;
    }
  }
  for (int i = 0; i < n; i++) {
    int row = i, col = 0;
    long long val = 0;
    while (check(row) and check(col)) {
      val += a[row][col];
      row--;
      col++;
    }
    row = i, col = 0;
    while (check(row) and check(col)) {
      diag[row][col] += val;
      diag[row][col] -= a[row][col];
      row--;
      col++;
    }
  }
  for (int i = 1; i < n; i++) {
    int row = n - 1, col = i;
    long long val = 0;
    while (check(row) and check(col)) {
      val += a[row][col];
      row--;
      col++;
    }
    row = n - 1, col = i;
    while (check(row) and check(col)) {
      diag[row][col] += val;
      diag[row][col] -= a[row][col];
      row--;
      col++;
    }
  }
  long long mx1 = 0, mx2 = 0;
  int r1 = 0, c1 = 0, r2 = 0, c2 = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (diag[i][j] >= mx2) {
          mx2 = diag[i][j];
          r2 = i;
          c2 = j;
        }
      } else {
        if (diag[i][j] >= mx1) {
          mx1 = diag[i][j];
          r1 = i;
          c1 = j;
        }
      }
    }
  }
  printf("%lld\n", mx1 + mx2);
  ;
  printf("%d %d %d %d\n", r1 + 1, c1 + 1, r2 + 1, c2 + 1);
}
