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>
#pragma comment(linker, "/STACK:102400000,1024000")
using namespace std;
long long num[2010][2010], ans[2][2010][2010];
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) scanf("%I64d", &num[i][j]);
  memset(ans, 0, sizeof(ans));
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j) ans[0][i][j] = ans[0][i - 1][j - 1] + num[i][j];
  for (i = n - 1; i >= 1; --i)
    for (j = n - 1; j >= 1; --j) ans[0][i][j] = ans[0][i + 1][j + 1];
  for (i = 1; i <= n; ++i)
    for (j = n; j >= 1; --j) ans[1][i][j] = ans[1][i - 1][j + 1] + num[i][j];
  for (i = n - 1; i >= 1; --i)
    for (j = 2; j <= n; ++j) ans[1][i][j] = ans[1][i + 1][j - 1];
  long long Max[2] = {0, 0}, x[2] = {0, 0}, y[2] = {0, 0};
  int ini = 0, sta;
  for (i = 1; i <= n; ++i) {
    sta = ini;
    for (j = 1; j <= n; ++j) {
      if (Max[sta] <= ans[0][i][j] + ans[1][i][j] - num[i][j])
        x[sta] = i, y[sta] = j,
        Max[sta] = ans[0][i][j] + ans[1][i][j] - num[i][j];
      sta ^= 1;
    }
    ini ^= 1;
  }
  printf("%I64d\n%I64d %I64d %I64d %I64d\n", Max[0] + Max[1], x[0], y[0], x[1],
         y[1]);
  return 0;
}
