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>
const int N = 2010;
int a[N][N];
long long sum[4][N][N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) scanf("%d", a[i] + j);
  for (int i = 1; i <= n; ++i)
    for (int j = n; j >= 1; --j) sum[0][i][j] = sum[0][i - 1][j + 1] + a[i][j];
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) sum[1][i][j] = sum[1][i - 1][j - 1] + a[i][j];
  for (int i = n; i >= 1; --i)
    for (int j = 1; j <= n; ++j) sum[2][i][j] = sum[2][i + 1][j - 1] + a[i][j];
  for (int i = n; i >= 1; --i)
    for (int j = n; j >= 1; --j) sum[3][i][j] = sum[3][i + 1][j + 1] + a[i][j];
  long long best1 = -1;
  int x1 = 0, y1 = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = (i & 1) + 1; j <= n; j += 2) {
      long long val = sum[0][i][j] + sum[1][i][j] + sum[2][i][j] +
                      sum[3][i][j] - 3LL * a[i][j];
      if (val > best1) {
        best1 = val;
        x1 = i;
        y1 = j;
      }
    }
  }
  long long best2 = -1;
  int x2 = 0, y2 = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = (!(i & 1)) + 1; j <= n; j += 2) {
      long long val = sum[0][i][j] + sum[1][i][j] + sum[2][i][j] +
                      sum[3][i][j] - 3LL * a[i][j];
      if (val > best2) {
        best2 = val;
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%I64d\n", best1 + best2);
  printf("%d %d %d %d", x1, y1, x2, y2);
}
