Kostya is a genial sculptor, he has an idea: to carve a marble sculpture in the shape of a sphere. Kostya has a friend Zahar who works at a career. Zahar knows about Kostya's idea and wants to present him a rectangular parallelepiped of marble from which he can carve the sphere. 

Zahar has *n* stones which are rectangular parallelepipeds. The edges sizes of the *i*-th of them are *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> and *c*<sub class="lower-index">*i*</sub>. He can take no more than two stones and present them to Kostya. 

If Zahar takes two stones, he should glue them together on one of the faces in order to get a new piece of rectangular parallelepiped of marble. Thus, it is possible to glue a pair of stones together if and only if two faces on which they are glued together match as rectangles. In such gluing it is allowed to rotate and flip the stones in any way. 

Help Zahar choose such a present so that Kostya can carve a sphere of the maximum possible volume and present it to Zahar.
The first line contains the integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>).

*n* lines follow, in the *i*-th of which there are three integers *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> and *c*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub>, *c*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>) — the lengths of edges of the *i*-th stone. Note, that two stones may have exactly the same sizes, but they still will be considered two different stones.
In the first example we can connect the pairs of stones:
 -  2 and 4, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 -  2 and 5, the size of the parallelepiped: 3 × 2 × 8 or 6 × 2 × 4 or 3 × 4 × 4, the radius of the inscribed sphere 1, or 1, or 1.5 respectively. -  2 and 6, the size of the parallelepiped: 3 × 5 × 4, the radius of the inscribed sphere 1.5 -  4 and 5, the size of the parallelepiped: 3 × 2 × 5, the radius of the inscribed sphere 1 -  5 and 6, the size of the parallelepiped: 3 × 4 × 5, the radius of the inscribed sphere 1.5 
Or take only one stone:
 -  1 the size of the parallelepiped: 5 × 5 × 5, the radius of the inscribed sphere 2.5 -  2 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 -  3 the size of the parallelepiped: 1 × 4 × 1, the radius of the inscribed sphere 0.5 -  4 the size of the parallelepiped: 2 × 1 × 3, the radius of the inscribed sphere 0.5 -  5 the size of the parallelepiped: 3 × 2 × 4, the radius of the inscribed sphere 1 -  6 the size of the parallelepiped: 3 × 3 × 4, the radius of the inscribed sphere 1.5 
It is most profitable to take only the first stone.
6
5 5 5
3 2 4
1 4 1
2 1 3
3 2 4
3 3 4
1
1
7
10 7 8
5 10 3
4 2 6
5 5 5
10 2 8
4 2 1
7 7 7
2
1 5
1
1 1 1
1
1
2
2 3 1
2 2 3
2
2 1
1
1000000000 1000000000 1000000000
1
1
3
100 100 100
25 63 11
63 15 11
1
1
2
999999999 1000000000 1000000000
1000000000 1000000000 1000000000
2
2 1
3
1 1 2
1 2 2
1 1 1
1
1
3
500 1000 1000
1000 499 1000
999 999 999
2
1 2
3
500 1000 1000
1000 499 1000
1000 1001 1001
1
3
9
1 3 2
3 3 1
3 1 2
3 3 2
2 2 2
3 2 1
3 3 1
3 3 1
2 1 2
2
4 8
3
20 30 5
20 30 6
10 10 10
2
2 1
3
5 20 30
6 20 30
10 10 10
2
2 1
3
20 5 30
20 6 30
10 10 10
2
2 1
3
20 30 5
30 20 6
10 10 10
2
2 1
3
20 30 5
6 20 30
10 10 10
2
2 1
3
20 30 5
6 30 20
10 10 10
2
2 1
3
20 30 5
20 6 30
10 10 10
2
2 1
3
20 30 5
30 6 20
10 10 10
2
2 1
3
20 5 30
20 30 6
10 10 10
2
2 1
3
20 5 30
30 20 6
10 10 10
2
2 1
3
20 5 30
6 20 30
10 10 10
2
2 1
3
20 5 30
6 30 20
10 10 10
2
2 1
3
20 5 30
30 6 20
10 10 10
2
2 1
3
5 20 30
20 30 6
10 10 10
2
2 1
3
5 20 30
30 20 6
10 10 10
2
2 1
3
5 20 30
6 30 20
10 10 10
2
2 1
3
5 20 30
20 6 30
10 10 10
2
2 1
3
5 20 30
30 6 20
10 10 10
2
2 1
import random

def get_size(N):
    sizes = [23, 71, 151, 571, 1021, 1571, 5795,  11311, 15451, 50821, 96557, 160001, 262419]
    size = 11
    idx = 0
    while size < 2*N:
        size = sizes[idx]
        idx += 1
    return size

class Hash(object):
    def __init__(self, size):
        self.size = size
        self.items = [[] for i in range(size)]
        self.p1 = random.randint(1, size - 1)
        self.p2 = random.randint(1, size - 1)
        
    def count_index(self, b, c):
        return (b*self.p1 + c*self.p2)//self.size
    
    def get_pair(self, item):
        a, b, c, num = item
        idx = self.count_index(b, c)
        arr = self.items[idx]
        cur = None
        for i, cur in enumerate(arr):
            if cur[1] == b and cur[2] == c:
                break
            cur = None
        if cur:
            if cur[0] >= a:
                return cur[-1], item[-1], cur[0] + a
            else:
                self.items[idx][i] = item
                return item[-1], cur[-1], cur[0] + a
        else:
            self.items[idx].append(item)
            return None

N = int(input())
R_max = 0
best_one = None
best_pair = None
if_one = True
size = get_size(N)
hash = Hash(size)
for i in range(N):
    box = input().split()
    box = sorted([int(side) for side in box]) + [i + 1]
    if box[0] > R_max:
        best_one = i + 1
        R_max = box[0]
        if_one = True
    cur_pair = hash.get_pair(box)
    if cur_pair:
        box[0] = cur_pair[-1]
        R_paired = min(box[:-1])
        if R_paired > R_max:
            R_max = R_paired
            best_pair = cur_pair[:-1]
            if_one = False
            
#file = open('output.txt', 'w')
if if_one:
    #file.write(str(1) + '\n' + str(best_one) + '\n' + str(R_max))
    print(1)
    print(best_one)
    print(R_max)
else:
    #file.write(str(2) + '\n' + str(best_pair[0]) + ' ' + str(best_pair[1]) + '\n' + str(R_max))
    print(2)
    print(str(best_pair[0]) + ' ' + str(best_pair[1]))
    print(R_max)