Educational Codeforces Round 80 (Rated for Div. 2) D. Minimax Problem

D. Minimax Problem

time limit per test

5 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

You are given nn arrays a1a1, a2a2, ..., anan; each array consists of exactly mm integers. We denote the yy-th element of the xx-th array as ax,yax,y.

You have to choose two arrays aiai and ajaj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j). After that, you will obtain a new array bb consisting of mmintegers, such that for every k∈[1,m]k∈[1,m] bk=max(ai,k,aj,k)bk=max(ai,k,aj,k).

Your goal is to choose ii and jj so that the value of mink=1mbkmink=1mbk is maximum possible.

Input

The first line contains two integers nn and mm (1≤n≤3⋅1051≤n≤3⋅105, 1≤m≤81≤m≤8) — the number of arrays and the number of elements in each array, respectively.

Then nn lines follow, the xx-th line contains the array axax represented by mm integers ax,1ax,1, ax,2ax,2, ..., ax,max,m (0≤ax,y≤1090≤ax,y≤109).

Output

Print two integers ii and jj (1≤i,j≤n1≤i,j≤n, it is possible that i=ji=j) — the indices of the two arrays you have to choose so that the value of mink=1mbkmink=1mbk is maximum possible. If there are multiple answers, print any of them.

Example

input

Copy

6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0

output

Copy

1 5

/*
看到题目首先想到的就是二分答案，那么找到一个可能的答案之后就可以把所有数列转化成01数列，1代表该位置的数大于等于答案，0代表该位置的数小于答案。

选择两个数列使其满足答案，也就是使这两个数列的或和为（1<<m）-1，很显然不能n*n的枚举每个数列，但是2^8只有256，所以可以直接枚举数列对应的二进制数，判断答案是否可行，如果可行，就可以O（n）地寻找可行解
*/
#include<iostream>
#include<cstdio>
#include<bitset>
#include<cstring>
using namespace std;
int n,m,a[300010][10],ans1,ans2;
int b[300010];
bool vis[300];
bool check(int x){
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(a[i][j]>=x)b[i]^=1<<(j-1);
        }
    }
    memset(vis,0,sizeof(vis));
    for(int i=1;i<=n;i++)vis[b[i]]=1;
    bool flag=0;
    int s1=0,s2=0;
    for(int sta1=0;sta1<(1<<m);sta1++){
        for(int sta2=sta1;sta2<(1<<m);sta2++){
            if((sta1|sta2)==(1<<m)-1&&vis[sta1]&&vis[sta2]){
                s1=sta1;s2=sta2;
                flag=1;
                break;
            }
        }
        if(flag)break;
    }
    if(!flag)return 0;
    for(int i=1;i<=n;i++){
        if(b[i]==s1){
            ans1=i;
            break;
        }
    }
    for(int i=1;i<=n;i++){
        if(b[i]==s2){
            ans2=i;
            break;
        }
    }
    return 1;
}
int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++){
            scanf("%d",&a[i][j]);
        }
    int l=0,r=1000000000;
    while(l<=r){
        int mid=(l+r)>>1;
        if(check(mid)){
            l=mid+1;
        }
        else r=mid-1;
    }
    printf("%d %d\n",ans1,ans2);
    return 0;
}