#include <bits/stdc++.h>
using namespace std;
long long save[2005][2005];
long long dp1[2005][2005];
long long dp2[2005][2005];
struct node {
  int x, y;
  long long val;
  node() {}
  node(int _x, int _y, long long _val) : x(_x), y(_y), val(_val) {}
};
vector<node> cq[2];
int n;
bool ok(int x, int y) {
  if (x <= n && x >= 1 && y <= n && y >= 1) return true;
  return false;
}
long long dfs(int x, int y, long long sum) {
  if (!ok(x, y)) return sum;
  dp1[x][y] = dfs(x + 1, y + 1, sum + save[x][y]);
  return dp1[x][y];
}
long long Dfs(int x, int y, long long sum) {
  if (!ok(x, y)) return sum;
  dp2[x][y] = Dfs(x + 1, y - 1, sum + save[x][y]);
  return dp2[x][y];
}
int main() {
  scanf("%d", &n);
  cq[0].clear();
  cq[1].clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) scanf("%I64d", &save[i][j]);
  }
  memset(dp1, -1, sizeof dp1);
  memset(dp2, -1, sizeof dp2);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (dp1[i][j] < 0) {
        dfs(i, j, 0);
      }
      if (dp2[i][j] < 0) {
        Dfs(i, j, 0);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cq[(i + j) % 2].push_back(node(i, j, dp1[i][j] + dp2[i][j] - save[i][j]));
    }
  }
  node ans1;
  ans1.val = -1;
  for (int i = 0; i < cq[0].size(); i++) {
    if (ans1.val < cq[0][i].val) {
      ans1 = cq[0][i];
    }
  }
  node ans2;
  ans2.val = -1;
  for (int i = 0; i < cq[1].size(); i++) {
    if (ans2.val < cq[1][i].val) {
      ans2 = cq[1][i];
    }
  }
  printf("%I64d\n", ans1.val + ans2.val);
  printf("%d %d %d %d\n", ans1.x, ans1.y, ans2.x, ans2.y);
  return 0;
}

### Prompt

Create a solution in cpp for 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 save[2005][2005];
long long dp1[2005][2005];
long long dp2[2005][2005];
struct node {
  int x, y;
  long long val;
  node() {}
  node(int _x, int _y, long long _val) : x(_x), y(_y), val(_val) {}
};
vector<node> cq[2];
int n;
bool ok(int x, int y) {
  if (x <= n && x >= 1 && y <= n && y >= 1) return true;
  return false;
}
long long dfs(int x, int y, long long sum) {
  if (!ok(x, y)) return sum;
  dp1[x][y] = dfs(x + 1, y + 1, sum + save[x][y]);
  return dp1[x][y];
}
long long Dfs(int x, int y, long long sum) {
  if (!ok(x, y)) return sum;
  dp2[x][y] = Dfs(x + 1, y - 1, sum + save[x][y]);
  return dp2[x][y];
}
int main() {
  scanf("%d", &n);
  cq[0].clear();
  cq[1].clear();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) scanf("%I64d", &save[i][j]);
  }
  memset(dp1, -1, sizeof dp1);
  memset(dp2, -1, sizeof dp2);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (dp1[i][j] < 0) {
        dfs(i, j, 0);
      }
      if (dp2[i][j] < 0) {
        Dfs(i, j, 0);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      cq[(i + j) % 2].push_back(node(i, j, dp1[i][j] + dp2[i][j] - save[i][j]));
    }
  }
  node ans1;
  ans1.val = -1;
  for (int i = 0; i < cq[0].size(); i++) {
    if (ans1.val < cq[0][i].val) {
      ans1 = cq[0][i];
    }
  }
  node ans2;
  ans2.val = -1;
  for (int i = 0; i < cq[1].size(); i++) {
    if (ans2.val < cq[1][i].val) {
      ans2 = cq[1][i];
    }
  }
  printf("%I64d\n", ans1.val + ans2.val);
  printf("%d %d %d %d\n", ans1.x, ans1.y, ans2.x, ans2.y);
  return 0;
}

```