#include <iostream>

// Greedy O(n^3) solution to diploma ceremony
// Author: Torstein Stromme

using namespace std;

#define MAXN 5000
#define rep(i,n) for(int i=0;i<(n);i++)

int graph[MAXN][MAXN], is_winner[MAXN], a, b, n, worst_k, running_k;

int main() {
    // Read graph
    scanf("%d", &n);
    rep(i, n-1) rep(j, i+1) {
        char c;
        scanf(" %c", &c);
        if (c == '1') {
            graph[i+1][j] = 1;
        }
        else {
            graph[j][i+1] = 1;
        }
    }
    
    // Greedily make a winner of the one who has smallest "new_bad-old_bad"
    rep(ignore, n) {
        int best_score = n + 1;
        int best_id = n + 1;
        rep(i, n) if (!is_winner[i]) {
            // Calculate this person's score
            int score = 0;
            rep(j, n) if (i != j) {
                if (is_winner[j] && graph[i][j]) score--;
                if (!is_winner[j] && !graph[i][j]) score++;
            }
            
            // Remember him if he is the best so far
            best_id = score < best_score ? i : best_id;
            best_score = score < best_score ? score : best_score;
        }
        is_winner[best_id] = 1;
        running_k += best_score;
        worst_k = running_k > worst_k ? running_k : worst_k;
    }

    cout << worst_k << endl;
    return 0;
}