Codeforces Round #378 (Div. 2)D（MAP）

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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：给你n个长方体，可以选一个，也可以选择俩拼在一起（至少有一面相同），问使得内接球体积最大的长方体的下标。
题解：
内接球的体积跟半径正相关，半径跟所选长方体的最小边正相关，我们要求的就是最长的最小边。可以贪心，也可以map暴力，这里用的是后者.在输入时先求出一个长方形的最大可能，把(a（长）,b（宽）,c（高）,d（编号）)保存到map里。（三个面，每个放入一次），再询问是否有能拼接的长方体，求出此时的最短边。如果放入的长方形比更优，更新map。
代码：

#include <bits/stdc++.h>#include <stdio.h>using namespace std;typedef pair<int,int>p;map<p,p>mp;int n,ans,pos,pos1,flag;void add(int a,int b,int c,int d){    if(mp.find(p(a,b))==mp.end())        mp[p(a,b)]=make_pair(c,d);    else    {        p l=mp[p(a,b)];        int t=l.first;        int tmp=min(min(a,b),t+c);        if(tmp>ans)        {            pos=d;            pos1=l.second;            ans=tmp;            flag=2;        }        if(t<c)        mp[p(a,b)]=make_pair(c,d);    }}int main(){    mp.clear();    scanf("%d",&n);    ans=pos=pos1=flag=0;    for(int i=1;i<=n;i++)    {        int w[4];        scanf("%d%d%d",&w[1],&w[2],&w[3]);        sort(w+1,w+4);        if(w[1]>ans)        {            pos=i;            flag=1;            ans=w[1];        }        add(w[1],w[2],w[3],i);        add(w[2],w[3],w[1],i);        add(w[1],w[3],w[2],i);    }    printf("%d\n",flag);    if(flag==1)    {        printf("%d\n",pos);    }    else    {        if(pos>pos1)            swap(pos,pos1);        printf("%d %d\n",pos,pos1);    }    return 0;}