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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long N = 2e3 + 10;
const long long inf = 1e18 + 10;
long long n, a[N][N], cheo1[2 * N], cheo2[2 * N], ma[2], x[2], y[2];
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cin >> a[i][j];
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cheo1[i + j] += a[i][j];
      cheo2[i - j + n] += a[i][j];
    }
  }
  for (long long i = 0; i <= 1; i++) {
    ma[i] = -inf;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      long long sum = cheo1[i + j] + cheo2[i - j + n] - a[i][j];
      long long t = (i + j) % 2;
      if (sum > ma[t]) {
        x[t] = i;
        y[t] = j;
        ma[t] = sum;
      }
    }
  }
  cout << ma[0] + ma[1] << "\n";
  cout << x[0] << " " << y[0] << " " << x[1] << " " << y[1];
}
