#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define is_mul_overflow(a, b) \ ((b != 0) && (a > LLONG_MAX / b || a < LLONG_MIN / b)) typedef pair pli; typedef pair pil; typedef pair pll; typedef pair pii; typedef pair pdd; typedef pair piii; typedef pair pil; typedef pair plii; typedef pair pdi; typedef long long ll; typedef unsigned long long ull; typedef pair puu; typedef long double ld; const int N = 2000086, MOD = 998244353, INF = 0x3f3f3f3f, MID = 333; const long double EPS = 1e-8; int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1}; // int dx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy[8] = {0, 1, 1, 1, 0, -1, -1, -1}; // int dx[8] = {2, 1, -1, -2, -2, -1, 1, 2}, dy[8] = {1, 2, 2, 1, -1, -2, -2, -1}; int n, m, cnt; int w[N]; vector num; ll res; ll lowbit(ll x) { return x & -x; } ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } ll lcm(ll a, ll b) { return a / gcd(a, b) * b; } inline double rand(double l, double r) { return (double)rand() / RAND_MAX * (r - l) + l; } inline ll qmi(ll a, ll b, ll c) { ll res = 1; while (b) { if (b & 1) res = res * a % c; a = a * a % c; b >>= 1; } return res; } inline ll qmi(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } inline double qmi(double a, ll b) { double res = 1; while (b) { if (b & 1) res *= a; a *= a; b >>= 1; } return res; } // inline ll C(ll a, ll b) { if (a < b) return 0; if (b > a - b) b = a - b; ll res = 1; for (ll i = 1, j = a; i <= b; i++, j--) { res = res * (j % MOD) % MOD; res = res * qmi(i, MOD - 2, MOD) % MOD; } return res; } inline ll C(ll a, ll b, int* c) { if (a < b) return 0; ll res = 1; for (ll j = a, i = 1; i < b + 1; i++, j--) res *= j; for (ll j = a, i = 1; i < b + 1; i++, j--) res /= i; return res; } inline int find_(int x) { return lower_bound(num.begin(), num.end(), x) - num.begin(); } int a[5086][3]; ll f[2][5086][3]; inline void solve() { memset(f[0], -0x3f, sizeof f[0]); f[0][0][0] = f[0][0][1] = f[0][0][2] = 0; for (int i = 1; i < n + 1; i++) { memset(f[i & 1], -0x3f, sizeof f[i & 1]); for (int j = 0; j <= n; j++) for (int k = 0; k < 3; k++) for (int l = 0; l < 3; l++) if (k != l) { int lv = max(0, j - 1) + 1, rv = min(n - i, lv - 1 + a[i][l]); ll t = 0; if (lv - 1 + a[i][l] > n - i) t = (ll)min(a[i][l], lv - 1 + a[i][l] - (n - i)) * (n - i + 1); if (rv >= lv) t += (ll)(lv + rv) * (rv - lv + 1) / 2; f[i & 1][min(n, max(0, j - 1) + a[i][l])][l] = max(f[i & 1][min(n, max(0, j - 1) + a[i][l])][l], f[i & 1 ^ 1][j][k] + t); } } for (int i = 0; i < n + 1; i++) res = max({res, f[n & 1][i][0], f[n & 1][i][1], f[n & 1][i][2]}); printf("%lld\n", res); } int main() { cin >> n; for (int i = 1; i < n + 1; i++) scanf("%d%d%d", a[i], a[i] + 1, a[i] + 2); solve(); return 0; }