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
from sys import stdin, stdout

n = int(stdin.readline())
ls = []
for i in range(n):
	ls.append(list(map(int, stdin.readline().split())))
mov1 = [0] * (2 * n - 1)
mov2 = [0] * (2 * n - 1)

for i in range(n):
	for j in range(n):
		mov1[i - j + n - 1] += ls[i][j]
		mov2[i + j] += ls[i][j]
mx1 = -1
p1 = [1, 1]
mx2 = -1
p2 = [1, 1]
for i in range(n):
	for j in range(n):
		if (i + j) % 2 == 0:
			if mov1[i - j + n - 1] + mov2[i + j] - ls[i][j] > mx1:
				mx1 = mov1[i - j + n - 1] + mov2[i + j] - ls[i][j]
				p1 = [i + 1, j + 1]
		else:
			if mov1[i - j + n - 1] + mov2[i + j] - ls[i][j] > mx2:
				mx2 = mov1[i - j + n - 1] + mov2[i + j] - ls[i][j]
				p2 = [i + 1, j + 1]

ans = str(mx1 + mx2) + '\n' + str(p1[0]) + ' ' + str(p1[1]) + ' ' + str(p2[0]) + ' ' + str(p2[1])
stdout.write(ans)
