#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
using namespace std;int N, G[25][25];int cal(int x) {int A[25], B[25], ca = 0, cb = 0, ret = 0;for (int i = 0; i < N; ++i) {if (x & (1 << i)) {A[ca++] = i + 1;} else {B[cb++] = i + 1;    }}for (int i = 0; i < ca; ++i) {for (int j = 0; j < cb; ++j) {ret += G[A[i]][B[j]];}    }return ret;
}int main() {int ret;while (scanf("%d", &N) == 1) {ret = 0;for (int i = 1; i <= N; ++i) {for (int j = 1; j <= N; ++j) {scanf("%d", &G[i][j]);}} // 对给定信息进行存储int mask = (1 << N) - 1;for (int i = 1; i < mask; ++i) {ret = max(ret, cal(i));}printf("%d\n", ret);}return 0;
}

#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <ctime>
#include <map>
using namespace std;
// 由于问题的规模相对较小
// 可以使用随机化算法来解决这个问题 int N, G[30][30];int randtime = 200000;int A[30];void MoveAToB(int x, int &t) {A[x] = 0;for (int i = 1; i <= N; ++i) {if (A[i]) t += G[x][i];else t -= G[x][i];}
}void MoveBToA(int x, int &t) {A[x] = 1;for (int i = 1; i <= N; ++i) {if (!A[i]) t += G[x][i];else t -= G[x][i];}
}int randpro() {int rd, ret = 0, t = 0;for (int i = 0; i < randtime; ++i) {rd = rand() % N + 1; // 得到一个点，这个点被if (A[rd]) { // 说明在A集合
            MoveAToB(rd, t);} else {  // 说明在B集合
            MoveBToA(rd, t);}ret = max(ret, t);}return ret;
}int main() {srand(time(NULL));while (scanf("%d", &N) == 1) {for (int i = 1; i <= N; ++i) {for (int j = 1; j <= N; ++j) {scanf("%d", &G[i][j]);}}for (int i = 1; i <= N; ++i) {A[i] = 1; // 先假设所有的元素都是属于第一个集合
        }printf("%d\n", randpro());}    return 0;
}