Codeforces_733D

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 ai, bi and ci. 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 ai, bi and ci (1 ≤ ai, bi, ci ≤ 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：给定n个长方体的长宽高，一次可以取一个，或者如果两个长方体有相同面，那么可以将这两个长方体通过相同面连接起来，最多可以连两个长方体。问要取的长方体的内接球的半径最大，应该取哪一个或者哪两个。

 

思路：先算出取一个能取的最大值，再算两个的。长方体内接球的半径受限于最短边，也就是说如果要将两个拼接起来，若不增加最短边，那么结果不可能大于取一个的最大值。接下来就暴力了。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<set>
using namespace std;
#define N 100005
#define INF 1e9+5
struct Node
{
    int a,b,c,index;
} node[N];

bool cmp(Node x,Node y)
{
    if(x.a>y.a)
        return 1;
    else if(x.a==y.a)
    {
        if(x.b>y.b)
            return 1;
        else if(x.b==y.b)
        {
            if(x.c>y.c)
                return 1;
            else
                return 0;
        }
        else
            return 0;
    }
    else
        return 0;
}

int getR(int a,int b,int c)
{
    int minx=INF;
    minx=min(minx,a);
    minx=min(minx,b);
    minx=min(minx,c);
    return minx;
}

int main()
{
    int n;
    scanf("%d",&n);
    int num[3];
    for(int i=0; i<n; i++)
    {
        scanf("%d%d%d",&num[0],&num[1],&num[2]);
        sort(num,num+3);
        node[i].a=num[2];
        node[i].b=num[1];
        node[i].c=num[0];
        node[i].index=i+1;
    }
    sort(node,node+n,cmp);
    int aa=-1,bb=-1,maxn=0;
    for(int i=0; i<n; i++)
    {
        int tmp=getR(node[i].a,node[i].b,node[i].c);
        if(tmp>maxn)
        {
            maxn=tmp;
            aa=node[i].index;
        }
    }
    for(int i=1; i<n; i++)
    {
        if(node[i].a==node[i-1].a&&node[i].b==node[i-1].b)
        {
            int tmp=getR(node[i].a,node[i].b,node[i].c+node[i-1].c);
            if(tmp>maxn)
            {
                maxn=tmp;
                aa=node[i].index;
                bb=node[i-1].index;
            }
        }
    }
    if(aa>0&&bb>0)
    {
        printf("2\n");
        printf("%d %d\n",aa,bb);
    }
    else
    {
        printf("1\n");
        printf("%d",aa);
    }
    return 0;
}