#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long powm(long long base, long long exp, long long mod = 1000000007) {
  base %= mod;
  long long ans = 1;
  while (exp) {
    if (exp & 1LL) ans = (ans * base) % mod;
    exp >>= 1LL, base = (base * base) % mod;
  }
  return ans;
}
long long a[2002][2002];
map<long long, long long> d1, d2;
long long b[2002][2002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  }
  for (int i = 1 - n; i <= n - 1; i++) {
    int sx, sy;
    if (i <= 0) {
      sx = 1;
      sy = 1 - i;
    } else {
      sx = 1 + i;
      sy = 1;
    }
    long long k = 0;
    for (int j = 0; sx + j <= n && sy + j <= n; j++) {
      k += a[sx + j][sy + j];
    }
    d1[i] = k;
  }
  for (int i = 2; i <= 2 * n; i++) {
    int sx, sy;
    if (i <= 1 + n) {
      sx = 1;
      sy = i - 1;
    } else {
      sx = i - n;
      sy = n;
    }
    long long k = 0;
    for (int j = 0; sx + j <= n && sy - j > 0; j++) {
      k += a[sx + j][sy - j];
    }
    d2[i] = k;
  }
  long long omax = 0, emax = 0;
  long long ox, oy, ex, ey;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long k = d1[i - j] + d2[i + j] - a[i][j];
      if (((i + j) & 1) && k >= omax) {
        omax = k;
        ox = i;
        oy = j;
      } else if (!((i + j) & 1) && k >= emax) {
        emax = k;
        ex = i;
        ey = j;
      }
    }
  }
  cout << omax + emax << '\n';
  cout << ex << " " << ey << " " << ox << " " << oy << '\n';
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long powm(long long base, long long exp, long long mod = 1000000007) {
  base %= mod;
  long long ans = 1;
  while (exp) {
    if (exp & 1LL) ans = (ans * base) % mod;
    exp >>= 1LL, base = (base * base) % mod;
  }
  return ans;
}
long long a[2002][2002];
map<long long, long long> d1, d2;
long long b[2002][2002];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) cin >> a[i][j];
  }
  for (int i = 1 - n; i <= n - 1; i++) {
    int sx, sy;
    if (i <= 0) {
      sx = 1;
      sy = 1 - i;
    } else {
      sx = 1 + i;
      sy = 1;
    }
    long long k = 0;
    for (int j = 0; sx + j <= n && sy + j <= n; j++) {
      k += a[sx + j][sy + j];
    }
    d1[i] = k;
  }
  for (int i = 2; i <= 2 * n; i++) {
    int sx, sy;
    if (i <= 1 + n) {
      sx = 1;
      sy = i - 1;
    } else {
      sx = i - n;
      sy = n;
    }
    long long k = 0;
    for (int j = 0; sx + j <= n && sy - j > 0; j++) {
      k += a[sx + j][sy - j];
    }
    d2[i] = k;
  }
  long long omax = 0, emax = 0;
  long long ox, oy, ex, ey;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      long long k = d1[i - j] + d2[i + j] - a[i][j];
      if (((i + j) & 1) && k >= omax) {
        omax = k;
        ox = i;
        oy = j;
      } else if (!((i + j) & 1) && k >= emax) {
        emax = k;
        ex = i;
        ey = j;
      }
    }
  }
  cout << omax + emax << '\n';
  cout << ex << " " << ey << " " << ox << " " << oy << '\n';
}

```