Kostya the Sculptor CodeForces

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 thei-th of them are a_i,b_i andc_i. He can takeno 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 andc_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 printk distinct integers from 1 to n — the numbers of stones which Zahar needs to choose. Consider that stones are numbered from1 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.

Example

Input

65 5 53 2 41 4 12 1 33 2 43 3 4

Output

11

Input

710 7 85 10 34 2 65 5 510 2 84 2 17 7 7

Output

21 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 or3 × 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.

当时没敢做， 主要是害怕超时，我们要是遍历的话，稍微一弄就超时了，所以果断放弃。

所以在博客上看到了两种做法，一种是用map,直接进行查询，这样就可以直接找到对应的bc，我们就可以进行比较。

另一种就是巧妙的处理数据，巧妙地排序，即a  b c,按照 b c,优先大来进行排序，这样的话，我们就可以找到两个相同的b c,那么a就可以增长了。

我们想 a b c; A B C,为了扩大最小的边，我们只能拿b c，去当做共同面，那么我们会和AB重合么，那么A<C  自然A<C+a,所以说最小的还是A，但是之前我们已经记录了单个的时候，所以这种情况可以忽略。那么A C，也是同理，所以只有BC，才有可能扩大最小边。

map

#include<iostream>#include<cstdio>#include <map>#include <string>#include<algorithm>using namespace std;typedef long long ll;typedef pair<ll,ll> pll;map<pll,pll> mp;int main(){      int n;      scanf("%d",&n);      ll max_len=0;      ll max_id,max_left,max_right;      int flag=0;      for(int i=0;i<n;i++){        ll val[3];        for(int j=0;j<3;j++)            scanf("%I64d",&val[j]);        sort(val,val+3);        if(val[0]>max_len){            max_len=val[0];            max_id=i+1;            flag=1;        }        pll tmp=pll(val[2],val[1]);        ll b[3];        b[2]=val[2],b[1]=val[1],b[0]=mp[tmp].first+val[0];        sort(b,b+3);        if(b[0]>max_len){            flag=0;            max_len=b[0];            max_left=mp[tmp].second;            max_right=i+1;        }        if(val[0]>mp[tmp].first){            mp[tmp].first=val[0];            mp[tmp].second=i+1;        }      }      if(!flag){        printf("2\n");        printf("%I64d %I64d\n",max_left,max_right);      }      else{        printf("1\n");        printf("%I64d\n",max_id);      }} 

巧妙处理

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;typedef long long ll;const int maxn=1e5;struct stone{    ll a,b,c;    int id;    bool operator<(const stone other)const{        if(c!=other.c)            return c<other.c;        if(b!=other.b)            return b<other.b;        return a<other.a;    }}s[maxn];int cmp(ll a,ll b){    return a<b;}int main(){   int n;   scanf("%d",&n);   ll max_ans=0,max_id;   for(int i=0;i<n;i++){       ll val[3];       for(int j=0;j<3;j++){        scanf("%I64d",&val[j]);       }       sort(val,val+3,cmp);       if(val[0]>max_ans){            max_ans=val[0];            max_id=i;       }      s[i].a=val[0],s[i].b=val[1],s[i].c=val[2];      s[i].id=i+1;   }   ll temp=max_ans,ans_left,ans_right;   sort(s,s+n);  // for(int i=0;i<n;i++)   // printf("%I64d %I64d %I64d\n",s[i].a,s[i].b,s[i].c);   for(int i=0;i<n-1;i++){    if(s[i].b==s[i+1].b&&s[i].c==s[i+1].c){            ll min_len=min(s[i].b,s[i].c);            min_len=min(min_len,s[i].a+s[i+1].a);            if(min_len>max_ans){                max_ans=min_len;                ans_left=s[i].id;                ans_right=s[i+1].id;            }        }   }  // printf("max:%I64d\n",max_ans);   if(max_ans==temp){      printf("1\n");      printf("%I64d\n",max_id+1);   }   else{     printf("2\n");     printf("%I64d %I64d\n",ans_left,ans_right);   }   return 0;}