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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 2e3 + 5;
const long long mod = 1e9 + 7;
int mp[maxn][maxn];
long long zhu[maxn << 1], fu[maxn << 1];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      scanf("%d", &mp[i][j]);
      zhu[i - j + maxn] += mp[i][j];
      fu[i + j] += mp[i][j];
    }
  int x1 = 0, y1 = 0;
  long long ans1 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = (i % 2 == 1) ? 1 : 2; j <= n; j += 2) {
      if (zhu[i - j + maxn] + fu[i + j] - mp[i][j] >= ans1) {
        ans1 = zhu[i - j + maxn] + fu[i + j] - mp[i][j];
        x1 = i;
        y1 = j;
      }
    }
  }
  int x2 = 0, y2 = 0;
  long long ans2 = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = (i % 2 == 1) ? 2 : 1; j <= n; j += 2) {
      if (zhu[i - j + maxn] + fu[i + j] - mp[i][j] >= ans2) {
        ans2 = zhu[i - j + maxn] + fu[i + j] - mp[i][j];
        x2 = i;
        y2 = j;
      }
    }
  }
  printf("%lld\n", ans1 + ans2);
  printf("%d %d %d %d", x1, y1, x2, y2);
}
