*cf.4 贪心

D. Kostya the Sculptor

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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_i, b_i and c_i. 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.

Input

The first line contains the integer n (1 ≤ n ≤ 10⁵).

n lines follow, in the i-th of which there are three integers a_i, b_i and c_i (1 ≤ a_i, b_i, c_i ≤ 10⁹) — 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.

Output

In the first line print k (1 ≤ k ≤ 2) the number of stones which Zahar has chosen. In the second line print k distinct integers from 1 to n — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from 1 to n in the order as they are given in the input data.

You can print the stones in arbitrary order. If there are several answers print any of them.

Examples

Input

6
5 5 5
3 2 4
1 4 1
2 1 3
3 2 4
3 3 4

Output

1
1

Input

7
10 7 8
5 10 3
4 2 6
5 5 5
10 2 8
4 2 1
7 7 7

Output

2
1 5

Note

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.

题意：

给出N个长方体，如果有两个长方体有两条边对应相等，即两个面的面积相同就可以把两个长方体粘起来，最多两个相粘，问最后那个或那两个可以有的最大内切球；

代码：

 //排序；普通的暴力会超时。后来看了别人的代码。神奇。
 //把每个长方体三条边从小到大排一下存入，以每个长方体最大的那条边从小到大排序如下。这样两个最大值和次大值对应相等的面必然相邻
 //这样每次比较相邻的两个就好了。如果最小的和第二大的对应相等怎么办如：(2 3 4),(1 2 5),(2 3 6)，输入数据的时候就排除了，就算合起来还是2，3 小边。
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 int n,flag1,flag2;
 struct cub
 {
     int a,b,c,ra;
 }cu[];
 bool cmp(cub x,cub y)           //排序
 {
     if(x.a==y.a&&x.b==y.b) return x.c<y.c;
     if(x.a==y.a) return x.b<y.b;
     return x.a<y.a;
 }
 int main()
 {
     int x,y,z;
     while(scanf("%d",&n)!=EOF)
     {
         int ans=;
         for(int i=;i<=n;i++)
         {
             scanf("%d%d%d",&x,&y,&z);
             int xx=max(x,max(y,z)),zz=min(x,min(y,z)),yy=x+y+z-xx-zz;
             cu[i].a=xx;
             cu[i].b=yy;
             cu[i].c=zz;
             cu[i].ra=i;
             if(zz>ans)
             {
                 ans=zz;
                 flag1=i;
                 flag2=;
             }
         }
         sort(cu+,cu++n,cmp);
         for(int i=;i<=n;i++)
         {
             if(cu[i].a==cu[i-].a&&cu[i].b==cu[i-].b)
             {
                 int Min=min(cu[i].c+cu[i-].c,min(cu[i].a,cu[i].b));
                 if(Min>ans)
                 {
                     ans=Min;
                     flag1=cu[i].ra;
                     flag2=cu[i-].ra;
                 }
             }
         }
         if(flag2==) printf("1\n%d\n",flag1);
         else printf("2\n%d %d\n",min(flag1,flag2),max(flag1,flag2));
     }
     return ;
 }