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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  long long arr[n][2005];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) cin >> arr[i][j];
  }
  long long sum[n][2005];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) sum[i][j] = 0;
  }
  for (int i = 0; i < n; i++) {
    long long temp = 0;
    int j = 0, x = i;
    while (x < n && j < n) {
      temp += arr[j][x];
      x++;
      j++;
    }
    j = 0;
    x = i;
    while (x < n && j < n) {
      sum[j][x] += temp;
      x++;
      j++;
    }
  }
  for (int i = 1; i < n; i++) {
    long long temp = 0;
    int j = 0, x = i;
    while (x < n && j < n) {
      temp += arr[x][j];
      x++;
      j++;
    }
    j = 0, x = i;
    while (x < n && j < n) {
      sum[x][j] += temp;
      x++;
      j++;
    }
  }
  for (int i = 0; i < n; i++) {
    long long temp = 0;
    int j = 0, x = i;
    while (x >= 0 && j >= 0 && j < n) {
      temp += arr[j][x];
      x--;
      j++;
    }
    j = 0, x = i;
    while (x >= 0 && j >= 0 && j < n) {
      sum[j][x] += temp;
      j++;
      x--;
    }
  }
  for (int i = 1; i < n; i++) {
    long long temp = 0;
    int j = n - 1, x = i;
    while (x >= 0 && x < n && j >= 0 && j < n) {
      temp += arr[x][j];
      x++;
      j--;
    }
    j = n - 1, x = i;
    while (x >= 0 && x < n && j >= 0 && j < n) {
      sum[x][j] += temp;
      x++;
      j--;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      sum[i][j] -= arr[i][j];
    }
  }
  int X1, Y1, X2, Y2;
  long long maxm = -1;
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (i % 2) {
      for (int j = 1; j < n; j += 2) {
        if (sum[i][j] > maxm) {
          maxm = sum[i][j];
          X1 = i;
          Y1 = j;
        }
      }
    } else {
      for (int j = 0; j < n; j += 2) {
        if (sum[i][j] > maxm) {
          maxm = sum[i][j];
          X1 = i;
          Y1 = j;
        }
      }
    }
  }
  ans += maxm;
  maxm = -1;
  for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
      for (int j = 1; j < n; j += 2) {
        if (sum[i][j] > maxm) {
          maxm = sum[i][j];
          X2 = i;
          Y2 = j;
        }
      }
    } else {
      for (int j = 0; j < n; j += 2) {
        if (sum[i][j] > maxm) {
          maxm = sum[i][j];
          X2 = i;
          Y2 = j;
        }
      }
    }
  }
  ans += maxm;
  cout << ans << "\n";
  cout << X1 + 1 << " " << Y1 + 1 << " " << X2 + 1 << " " << Y2 + 1;
  return 0;
}
