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;
int i, j, k;
const int M = 2000 + 6;
long long a[M][M];
long long pref1[M][M], pref2[M][M];
long long valu[M][M];
void prnt(long long alu[M][M], int n) {
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) cout << alu[i][j] << " ";
    cout << "\n";
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) cin >> a[i][j];
  }
  for (k = 1; k <= n; k++) {
    for (i = k, j = 1; i <= n and j <= n; i++, j++) {
      pref1[i][j] = a[i][j] + pref1[i - 1][j - 1];
    }
  }
  for (k = 2; k <= n; k++) {
    for (i = 1, j = k; i <= n and j <= n; i++, j++) {
      pref1[i][j] = a[i][j] + pref1[i - 1][j - 1];
    }
  }
  for (k = n; k >= 1; k--) {
    for (i = 1, j = k; i <= n and j >= 1; i++, j--) {
      pref2[i][j] = a[i][j] + pref2[i - 1][j + 1];
    }
  }
  for (k = 2; k <= n; k++) {
    for (i = k, j = n; i <= n and j >= 1; i++, j--) {
      pref2[i][j] = a[i][j] + pref2[i - 1][j + 1];
    }
  }
  int edge = n;
  for (k = 1; k <= n; k++) {
    for (i = k, j = 1; i <= n and j <= n; i++, j++) {
      valu[i][j] += pref1[n][edge] - a[i][j];
    }
    edge--;
  }
  edge = n - 1;
  for (k = 2; k <= n; k++) {
    for (i = 1, j = k; i <= n and j <= n; i++, j++) {
      valu[i][j] += pref1[edge][n] - a[i][j];
    }
    edge--;
  }
  edge = n;
  for (k = n; k >= 1; k--) {
    for (i = 1, j = k; i <= n and j >= 1; i++, j--) {
      valu[i][j] += pref2[edge][1];
    }
    edge--;
  }
  edge = 2;
  for (k = 2; k <= n; k++) {
    for (i = k, j = n; i <= n and j >= 1; i++, j--) {
      valu[i][j] += pref2[n][edge];
    }
    edge++;
  }
  long long sum1 = 0, sum2 = 0;
  int x1, y1, x2, y2;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2) continue;
      if (valu[i][j] >= sum1) {
        x1 = i, y1 = j;
        sum1 = valu[i][j];
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if ((i + j) % 2 == 0) continue;
      if (valu[i][j] >= sum2) {
        x2 = i, y2 = j;
        sum2 = valu[i][j];
      }
    }
  }
  cout << sum1 + sum2 << "\n";
  cout << x1 << " " << y1 << " " << x2 << " " << y2;
}
