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 = 2006;
long long l[N + N], r[N + N], a[N][N], sum[3][N][3], n;
int main() {
  scanf("%I64d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%I64d", &a[i][j]);
      l[i - j + n] += a[i][j];
      r[i + j - 1] += a[i][j];
    }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      a[i][j] = l[i - j + n] + r[i + j - 1] - a[i][j];
      if (j & 1) {
        if (a[i][j] > sum[1][i][0]) {
          sum[1][i][0] = a[i][j];
          sum[1][i][1] = j;
        }
      } else {
        if (a[i][j] > sum[0][i][0]) {
          sum[0][i][0] = a[i][j];
          sum[0][i][1] = j;
        }
      }
    }
  long long ans = 0, xy[2][2];
  xy[0][0] = 1;
  xy[0][1] = 1;
  xy[1][0] = 1;
  xy[1][1] = 2;
  for (int i = 1; i <= n; i++) {
    long long now = sum[1][i][0];
    long long maxt = sum[0][i][0], k = i;
    for (int j = 1; j <= n; j++)
      if (sum[abs(i - j) & 1][j][0] > maxt) {
        maxt = sum[abs(i - j) & 1][j][0];
        k = j;
      }
    now += maxt;
    if (now > ans) {
      ans = now;
      xy[0][0] = i;
      xy[0][1] = sum[1][i][1];
      xy[1][0] = k;
      xy[1][1] = sum[abs(k - i) & 1][k][1];
    }
    now = sum[0][i][0];
    maxt = sum[1][i][0], k = i;
    for (int j = 1; j <= n; j++)
      if (sum[!(abs(i - j) & 1)][j][0] > maxt) {
        maxt = sum[!(abs(i - j) & 1)][j][0];
        k = j;
      }
    now += maxt;
    if (now > ans) {
      ans = now;
      xy[0][0] = i;
      xy[0][1] = sum[0][i][1];
      xy[1][0] = k;
      xy[1][1] = sum[!(abs(k - i) & 1)][k][1];
    }
  }
  printf("%I64d\n", ans);
  printf("%I64d %I64d %I64d %I64d\n", xy[0][0], xy[0][1], xy[1][0], xy[1][1]);
}
