#include<bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(3)
//#define int long long
int n,k;
long long x[100005][6];
int fa[100005];
int find(int x){
	if(fa[x] == x)return x;
	return fa[x] = find(fa[x]);
}void merge(int x,int y){
	fa[find(x)] = find(y);
}long long mx[100005];int mxpos[100005];
int col[100005],ok[100005];vector<int>c[100005];int cnt =0 ;
void Bhoumianwangle(){
	long long mxx1[55],mxx2[55],mxxpos1[55],mxxpos2[55];
	for(int i =0;i<=(1<<k);i++)mxx1[i] = mxx2[i] = -1000000000000;
	cnt =0 ;
	for(int i =0;i<=n;i++)col[i] = ok[i] = 0,mx[i] = 0,c[i].clear();
	for(int i =1;i<=n;i++){
		if(!ok[find(i)])ok[find(i)] = ++cnt;
		col[i] = ok[find(i)];c[col[i]].push_back(i);
	}for(int i =1;i<=n;i++){
		for(int j=0;j<(1<<k);j++){
			long long sum = 0;
			for(int l = 0;l<k;l++){
				if(j>>l&1)sum = sum+x[i][l];
				else sum = sum-x[i][l];
			}if(sum>=mxx1[j]){
				mxx1[j] = sum;mxxpos1[j] = i;
			}
		}
	}for(int i = 1;i<=n;i++){
		for(int j =0;j<(1<<k);j++){
			long long sum = 0;
			for(int l =0;l<k;l++){
				if(j>>l&1)sum = sum+x[i][l];
				else sum = sum-x[i][l];
			}if(sum>=mxx2[j] and col[i]!=col[mxxpos1[j]]){
				mxx2[j] = sum;mxxpos2[j] = i;
			}
		}
	}for(int i = 1;i<=cnt;i++){
		for(int xx:c[i]){
			for(int j =0;j<(1<<k);j++){
				long long sum =0 ;
				for(int l =0;l<k;l++){
					if(j>>l&1)sum = sum-x[xx][l];
					else sum = sum+x[xx][l];
				}if(col[mxxpos1[j]] == i){
					if(sum+mxx2[j]>=mx[i]){
						mx[i] = sum+mxx2[j];mxpos[i] = mxxpos2[j];
					}
				}else{
					if(sum+mxx1[j]>=mx[i]){
						mx[i] = sum+mxx1[j];mxpos[i] = mxxpos1[j];
					}
				}
			}
		}
	}
}
signed main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin >>n >> k;
	for(int i =1;i<=n;i++){
		for(int j =0;j<k;j++)cin >> x[i][j];
	}for(int i = 1;i<=n;i++)fa[i] = i;long long sum =0 ;
	while(1){
		Bhoumianwangle();
		for(int i = 1;i<=cnt;i++){
			if(find(c[i].back()) !=find(mxpos[i]))sum+=mx[i];
			merge(c[i].back(),mxpos[i]);
		}int ok = 0;
		for(int i= 1;i<=n;i++)if(fa[i] == i)ok++;
		//for(int i = 1;i<=cnt;i++)cout << mx[i] << " " << mxpos[i] << endl;
		if(ok == 1)break;
	}cout << sum << endl;
	return 0;
}
/*
考虑使用 B后面忘了 生成树算法。
问题变成求到一个点最远的点。
由于绝对值的性质，把所有大小关系枚举一遍显然很对。
复杂度 O(2^kn\log^2 n) 
这题挺有意思。 

牛魔，居然要支持删数，这复杂度看着不大好啊。 
牛魔，不用支持删数，维护次大值就可以了。
写题五分钟，卡常一小时。 
*/