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;
inline long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
inline long long int fast_pow(long long int a, long long int b) {
  long long int res = 1;
  a %= 1000000007;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % 1000000007;
    a = a * a % 1000000007;
  }
  return res;
}
inline bool isPrime(long long int a) {
  for (long long int i = 2; i * i <= a; i++)
    if (a % i == 0) return 0;
  return 1;
}
const int N = 2010;
long long int arr[N][N], diagonalSum[N][N];
int n, x_1 = 0, x_2 = 0, y_1 = 0, y_2 = 1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cin >> arr[i][j];
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    long long int sum = 0;
    for (int j = 0, k = i; j < (n - i); j++, k++) sum += arr[k][j];
    for (int j = 0, k = i; j < (n - i); j++, k++) diagonalSum[k][j] += sum;
  }
  for (int i = 1; i < n; i++) {
    long long int sum = 0;
    for (int j = i, k = 0; j < n; j++, k++) sum += arr[k][j];
    for (int j = i, k = 0; j < n; j++, k++) diagonalSum[k][j] += sum;
  }
  for (int i = 0; i < n; i++) {
    long long int sum = 0;
    for (int j = 0, k = i; j < n && k >= 0; j++, k--) sum += arr[j][k];
    for (int j = 0, k = i; j < n && k >= 0; j++, k--) diagonalSum[j][k] += sum;
  }
  for (int i = 1; i < n; i++) {
    long long int sum = 0;
    for (int j = i, k = n - 1; j < n && k >= 0; j++, k--) sum += arr[j][k];
    for (int j = i, k = n - 1; j < n && k >= 0; j++, k--)
      diagonalSum[j][k] += sum;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      diagonalSum[i][j] -= arr[i][j];
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (diagonalSum[i][j] > diagonalSum[x_1][y_1]) {
          x_1 = i;
          y_1 = j;
        }
      } else {
        if (diagonalSum[i][j] > diagonalSum[x_2][y_2]) {
          x_2 = i;
          y_2 = j;
        }
      }
    }
  }
  cout << diagonalSum[x_1][y_1] + diagonalSum[x_2][y_2] << '\n';
  cout << x_1 + 1 << ' ' << y_1 + 1 << ' ' << x_2 + 1 << ' ' << y_2 + 1 << '\n';
  return 0;
}
