/**
   Author : Kamonasish Roy (Bullet) 
   Key idea:
   First we need to store each position frequancy.
   Then 1 to n, we check i and i+1, max frequency character for each position.
   Just combine and find maximum possible answer.
**/
#include<bits/stdc++.h>
using namespace std;
const long long M=4e5,MOD=1e9+7;
typedef long long ll;


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    cin>>t;
    while(t--){
    	int n,m;
    	cin>>n>>m;
    	vector<string>p;
    	vector<vector<int>>occ(m+1,vector<int>(26,0));
    	vector<vector<int>>occ1(m+1,vector<int>(26,0));
        for(int i=1;i<=n;i++){
        	string s;
        	cin>>s;
        	p.push_back(s);
        	for(int j=0;j<m;j++){
        		occ[j][s[j]-'a']++;
        	}
        }
        int best=0;
        for(int i=n-1;i>=0;i--){
        	int cur_best=0;
        	for(int j=0;j<m;j++){
        		int L=0;
        	    int R=0;
        		for(int k=0;k<26;k++)L=max(L,occ[j][k]),R=max(R,occ1[j][k]);
        		cur_best+=L+R;
        		
        	}
        	best=max(best,cur_best);
        	for(int j=0;j<m;j++){
        		char ch=p[i][j];
        		occ[j][ch-'a']--;
        		occ1[j][ch-'a']++;
        	}
        }
        cout<<best<<"\n";
    	
    	
    	
    	
    	}
    	
    	   
    return 0;
 
}