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;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long **a = new long long *[n + 1];
  for (long long i = 1; i <= n; i++) {
    a[i] = new long long[n + 1];
  }
  pair<long long, long long> **a2 = new pair<long long, long long> *[n + 1];
  for (long long i = 1; i <= n; i++) {
    a2[i] = new pair<long long, long long>[n + 1];
  }
  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++) {
      a2[i][j].first = 0;
      ;
      a2[i][j].second = 0;
      ;
    }
  }
  long long b[2010] = {};
  for (long long i = 0; i < n; i++) {
    b[i] = 0;
  }
  long long t = 1;
  long long k = 0;
  while (t <= n) {
    long long x = t;
    long long y = 1;
    while (x <= n && y <= n) {
      b[k] = b[k] + a[x][y];
      a2[x][y].first = k;
      x++;
      y++;
    }
    t += 2;
    k++;
  }
  long long u = 3;
  while (u <= n) {
    long long x = 1;
    long long y = u;
    while (x <= n && y <= n) {
      b[k] += a[x][y];
      a2[x][y].first = k;
      x++;
      y++;
    }
    u += 2;
    k++;
  }
  long long d[2010] = {};
  for (long long i = 0; i < n; i++) {
    d[i] = 0;
  }
  t = 1;
  u = 1;
  k = 0;
  while (t <= n) {
    long long x = t;
    long long y = 1;
    while (x >= 1 && y <= n) {
      d[k] += a[x][y];
      a2[x][y].second = k;
      x--;
      y++;
    }
    t += 2;
    k++;
  }
  t = n;
  u = (n % 2) + 2;
  while (u <= n) {
    long long x = n;
    long long y = u;
    while (x >= 1 && y <= n) {
      d[k] += a[x][y];
      a2[x][y].second = k;
      x--;
      y++;
    }
    u += 2;
    k++;
  }
  long long maxa = -1;
  long long r = 1;
  long long c = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 2 - (i % 2); j <= n; j += 2) {
      long long w = a2[i][j].first;
      long long z = a2[i][j].second;
      long long ans = b[w] + d[z] - a[i][j];
      if (ans > maxa) {
        maxa = ans;
        r = i;
        c = j;
      }
    }
  }
  long long h = r;
  long long m = c;
  for (long long i = 0; i < n; i++) {
    b[i] = 0;
  }
  t = 2;
  k = 0;
  while (t <= n) {
    long long x = t;
    long long y = 1;
    while (x <= n && y <= n) {
      b[k] = b[k] + a[x][y];
      a2[x][y].first = k;
      x++;
      y++;
    }
    t += 2;
    k++;
  }
  u = 2;
  while (u <= n) {
    long long x = 1;
    long long y = u;
    while (x <= n && y <= n) {
      b[k] += a[x][y];
      a2[x][y].first = k;
      x++;
      y++;
    }
    u += 2;
    k++;
  }
  for (long long i = 0; i < n; i++) {
    d[i] = 0;
  }
  for (long long i = 0; i < n; i++) {
    d[i] = 0;
  }
  t = 2;
  k = 0;
  while (t <= n) {
    long long x = t;
    long long y = 1;
    while (x >= 1 && y <= n) {
      d[k] += a[x][y];
      a2[x][y].second = k;
      x--;
      y++;
    }
    t += 2;
    k++;
  }
  u = 3 - (n % 2);
  while (u <= n) {
    long long x = n;
    long long y = u;
    while (x >= 1 && y <= n) {
      d[k] += a[x][y];
      a2[x][y].second = k;
      x--;
      y++;
    }
    u += 2;
    k++;
  }
  long long maxa2 = -1;
  r = 1;
  c = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1 + (i % 2); j <= n; j += 2) {
      long long w = a2[i][j].first;
      long long z = a2[i][j].second;
      long long ans = b[w] + d[z] - a[i][j];
      if (ans > maxa2) {
        maxa2 = ans;
        r = i;
        c = j;
      }
    }
  }
  cout << maxa + maxa2 << endl;
  cout << h << " " << m << " ";
  cout << r << " " << c << " ";
}
