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;
const int maxn = 2000 + 2;
long long a[maxn][maxn], row[maxn * 2], col[maxn * 2];
struct pp {
  long long v;
  int x, y;
} s[maxn * maxn / 2], t[maxn * maxn / 2];
int cmp(const pp &a, const pp &b) { return a.v > b.v; }
int main() {
  int i, j, n;
  while (scanf("%d", &n) != EOF) {
    memset(row, 0, sizeof(row));
    memset(col, 0, sizeof(col));
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        scanf("%I64d", &a[i][j]);
        row[j - i + n] += a[i][j];
        col[j + i] += a[i][j];
      }
    }
    int cnt1 = 0, cnt2 = 0;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= n; j++) {
        if ((i + j) & 1) {
          s[++cnt1].x = i;
          s[cnt1].y = j;
          s[cnt1].v = row[j - i + n] + col[j + i] - a[i][j];
        } else {
          t[++cnt2].x = i;
          t[cnt2].y = j;
          t[cnt2].v = row[j - i + n] + col[j + i] - a[i][j];
        }
      }
    }
    sort(s + 1, s + cnt1 + 1, cmp);
    sort(t + 1, t + cnt2 + 1, cmp);
    printf("%I64d\n", s[1].v + t[1].v);
    printf("%d %d %d %d\n", s[1].x, s[1].y, t[1].x, t[1].y);
  }
  return 0;
}
