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;
inline void print(char pt) { printf("%c\n", pt); }
inline void print(int pt) { printf("%d\n", pt); }
inline void print(long long pt) { printf("%I64d\n", pt); }
inline void print(double pt) { printf("%.20f\n", pt); }
inline void print(char pt[]) { printf("%s\n", pt); }
inline void print() { printf("\n"); }
inline int scan(int &pt) { return scanf("%d", &pt); }
inline int scan(char &pt) { return scanf("%c", &pt); }
inline int scan(long long &pt) { return scanf("%I64d", &pt); }
inline int scan(double &pt) { return scanf("%lf", &pt); }
inline int scan(char pt[]) { return scanf("%s", pt); }
struct pii {
  long long a;
  int i;
  friend int operator<(const pii &a, const pii &b) { return a.a > b.a; }
};
using namespace std;
int a[3000][3000];
long long b[3000][3000];
int n;
long long ka[6000];
long long kb[6000];
int main() {
  scan(n);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      scan(a[i][j]);
      ka[i + j] += a[i][j];
      kb[3000 + i - j] += a[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      b[i][j] = ka[i + j] + kb[3000 + i - j] - a[i][j];
    }
  }
  long long maxa = -1;
  int ia = 0;
  int ja = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i % 2; j < n; j += 2) {
      if (b[i][j] > maxa) {
        maxa = b[i][j];
        ia = i;
        ja = j;
      }
    }
  }
  long long maxb = -1;
  int ib = 0;
  int jb = 0;
  for (int i = 0; i < n; i++) {
    for (int j = !(i % 2); j < n; j += 2) {
      if (b[i][j] > maxb) {
        maxb = b[i][j];
        ib = i;
        jb = j;
      }
    }
  }
  print(maxa + maxb);
  print(ia + 1);
  print(ja + 1);
  print(ib + 1);
  print(jb + 1);
  return 0;
}
