n = int(input())
info = [list(map(int, input().split())) for i in range(n)]

# dp[bit_state][i][j] := bit_stateの積み木を使って、最後に使ったのがi番目の積み木。
#                        また、i番目の積み木のj番目の辺が高さになっている

# dp = [[[-1] * (3) for i in range(n)] for j in range(1 << n)]
dq = [[[-1] * (1 << n) for i in range(n)] for j in range(3)]
for i in range(n):
    for j, val in enumerate(info[i]):
        # dp[1 << i][i][j] = val
        dq[j][i][1 << i] = val

for bit_state in range(1 << n):
    for i in range(n):
        if not((1 << i) & bit_state):
            continue
        # i番目はすでに積まれているので、dp[bit_state][i][?]から遷移出来る
        
        for j in range(n):
            if (1 << j) & bit_state:
                continue

            # j番目はまだ積まれていないので、dp[bit_state|j][j][?] <- dp[bit_state][i][?]と遷移出来る
            for ii in range(3):
                for jj in range(3):
                    ai, bi = info[i][(ii + 1) % 3], info[i][(ii + 2) % 3]
                    aj, bj = info[j][(jj + 1) % 3], info[j][(jj + 2) % 3]
                    # if dp[bit_state][i][ii] == -1:
                    if dq[ii][i][bit_state] == -1:
                        continue
                    if (aj <= ai and bj <= bi) or (aj <= bi and bj <= ai):
                        # dp[bit_state|(1 << j)][j][jj] = max(dp[bit_state][i][ii] + info[j][jj], dp[bit_state|(1 << j)][j][jj])
                        dq[jj][j][bit_state|(1 << j)] = max(dq[ii][i][bit_state] + info[j][jj], dq[jj][j][bit_state|(1 << j)])
ans = 0
for bit_state in range(1 << n):
    for i in range(n):
        for j in range(3):
            ans = max(ans, dq[j][i][bit_state])
print(ans)