#include <iostream>
#include <vector>
#include <string>
#include <cstring> // for memset
#include <algorithm>
using namespace std;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int T;
    cin >> T;
    while(T--){
        int n, m;
        cin >> n >> m;
        vector<string> rounds(n);
        for (int i = 0; i < n; i++){
            cin >> rounds[i];
        }
        vector<long long> total(n+1, 0LL);
 
        for (int col = 0; col < m; col++){
            vector<int> prefix(n+1, 0);
            int freq[26] = {0};
            int currentMax = 0;
            for (int i = 1; i <= n; i++){
                char c = rounds[i-1][col];
                int idx = c - 'a';
                freq[idx]++;
                currentMax = max(currentMax, freq[idx]);
                prefix[i] = currentMax;
            }
 
            // suffix[i] = maximum frequency (in this column) among rounds i..n.
            vector<int> suffix(n+2, 0);  // suffix[n+1] = 0 (empty)
            memset(freq, 0, sizeof(freq));
            currentMax = 0;
            for (int i = n; i >= 1; i--){
                char c = rounds[i-1][col];
                int idx = c - 'a';
                freq[idx]++;
                currentMax = max(currentMax, freq[idx]);
                suffix[i] = currentMax;
            }
 
            // For each possible partition k (0 <= k <= n):
            // - Rounds 1..k use the guess-string chosen to maximize matches in that column (prefix part).
            // - Rounds k+1..n use a (possibly new) guess-string (suffix part).
            // So, add prefix[k] + suffix[k+1] to the total contribution for this column.
            for (int k = 0; k <= n; k++){
                total[k] += prefix[k] + suffix[k+1];
            }
        }
 
        // The answer is the maximum total score we can get over all possible partitions.
        long long ans = 0;
        for (int k = 0; k <= n; k++){
            ans = max(ans, total[k]);
        }
 
        cout << ans << "\n";
    }
    return 0;
}