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
import math, collections, sys
input = sys.stdin.readline
n = int(input())
grid = []
for i in xrange(n):
    grid.append(list(map(int, input().split(" "))))
d = [0 for i in xrange(2*n-1)]
pd = [0 for i in xrange(2*n-1)]
for i in xrange(n):
    for j in xrange(n):
        d[i+j]+=grid[i][j]
        pd[i+n-1-j]+=grid[i][j]
odd = [0, 0]
even = [0, 0]
oddval = -1
evenval = -1
for i in xrange(n):
    for j in xrange(n):
        if (i+j)%2:
            if d[i+j]+pd[i+n-1-j] - grid[i][j] > oddval:
                oddval = d[i+j]+pd[i+n-1-j]-grid[i][j]
                odd = [i+1, j+1]
        else:
            if d[i+j]+pd[i+n-1-j] - grid[i][j] > evenval:
                evenval = d[i+j]+pd[i+n-1-j]-grid[i][j]
                even = [i+1, j+1]
print(oddval+evenval)
print odd[0], odd[1], even[0], even[1]