Problem:
Score : 300 points

Problem Statement

Joisino is planning to open a shop in a shopping street.

Each of the five weekdays is divided into two periods, the morning and the evening. For each of those ten periods, a shop must be either open during the whole period, or closed during the whole period. Naturally, a shop must be open during at least one of those periods.

There are already N stores in the street, numbered 1 through N.

You are given information of the business hours of those shops, F_{i,j,k}. If F_{i,j,k}=1, Shop i is open during Period k on Day j (this notation is explained below); if F_{i,j,k}=0, Shop i is closed during that period. Here, the days of the week are denoted as follows. Monday: Day 1, Tuesday: Day 2, Wednesday: Day 3, Thursday: Day 4, Friday: Day 5. Also, the morning is denoted as Period 1, and the afternoon is denoted as Period 2.

Let c_i be the number of periods during which both Shop i and Joisino's shop are open. Then, the profit of Joisino's shop will be P_{1,c_1}+P_{2,c_2}+...+P_{N,c_N}.

Find the maximum possible profit of Joisino's shop when she decides whether her shop is open during each period, making sure that it is open during at least one period.

Constraints

1≤N≤100

0≤F_{i,j,k}≤1

For every integer i such that 1≤i≤N, there exists at least one pair (j,k) such that F_{i,j,k}=1.

-10^7≤P_{i,j}≤10^7

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
F_{1,1,1} F_{1,1,2} ... F_{1,5,1} F_{1,5,2}
:
F_{N,1,1} F_{N,1,2} ... F_{N,5,1} F_{N,5,2}
P_{1,0} ... P_{1,10}
:
P_{N,0} ... P_{N,10}

Output

Print the maximum possible profit of Joisino's shop.

Sample Input 1

1
1 1 0 1 0 0 0 1 0 1
3 4 5 6 7 8 9 -2 -3 4 -2

Sample Output 1

8

If her shop is open only during the periods when Shop 1 is opened, the profit will be 8, which is the maximum possible profit.

Sample Input 2

2
1 1 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1
0 -2 -2 -2 -2 -2 -1 -1 -1 -1 -1

Sample Output 2

-2

Note that a shop must be open during at least one period, and the profit may be negative.

Sample Input 3

3
1 1 1 1 1 1 0 0 1 1
0 1 0 1 1 1 1 0 1 0
1 0 1 1 0 1 0 1 0 1
-8 6 -2 -8 -8 4 8 7 -6 2 2
-9 2 0 1 7 -5 0 -2 -6 5 5
6 -6 7 -9 6 -5 8 0 -9 -7 -7

Sample Output 3

23

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
//#include <mylib.h>
using namespace std;
//cin.sync_with_stdio(false);

#define FOR(i,a,b) for(int i = (a); i < (b); ++i)
#define FOR_EQ(i,a,b) for(int i = (a); i <= (b); ++i)
#define FOR_RE(i,a,b) for(int i = (a); i > (b); --i)
#define rep(i,n) FOR(i,0,n)
#define rep_eq(i,n) FOR_EQ(i,0,n)
#define rep_re(i,n) FOR_RE(i,n,0)

#define DEBUG_VARIABLE(variable) cout << #variable << ":" << variable << endl

#define WHITE 1//（未訪問）
#define GRAY 2//（訪問）未だに訪問していない頂点への辺を持っている。スタックに退避。
#define BLACK 3//（訪問完了）未訪問への頂点への辺を持たない
#define N 100
#define INFTY (1 << 21) //10^21
#define INF (1 << 9)
int M[N][N];//隣接行列
int color[N];

//1個上から時計周り
const int ddx[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const int ddy[8] = {1, 1, 0, -1, -1, -1, 0, 1};
//上右下左
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {-1, 0, 1, 0};
static const int NIL = -1;

typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> Vector;
typedef vector<Vector>	DVector;
int n;

void printArray(int *array, int);
void printVector(Vector &v);
void printDimention(DVector &dv);

int main(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n;
    int f[n][10];
    ll p[n][11];
    
    rep(i, n){
        rep(j, 10){
            cin >> f[i][j];
        }
    }
    rep(i, n){
        rep(j, 11){
            cin >> p[i][j];
        }
    }
    
    int cnt[n];
    
    ll res = -10000000, tmp;
    for(int bit = 0; bit < (1 << 10); bit++){
        if(bit == 0)    continue;
        memset(cnt, 0, sizeof(cnt));
        rep(j, 10){
            if(bit & (1 << j)){
                rep(i, n){
                    if(f[i][j]){
                        cnt[i]++;
                    }
                }
            }
        }
        
        tmp = 0;
        rep(i, n){
            tmp += p[i][cnt[i]];
        }
        res = max(tmp, res);
    }
    cout << res << endl;
}

//配列の表示
void printArray(int array[], int n){
    rep(i, n){
        if(i)   cout << " ";
        cout << array[i];
    }
    cout << endl;
}

//Vectorの表示
void printVector(Vector &v){
    rep(i, (int)v.size()){
        if(i)   cout << " ";
        cout << v[i];
    }
    cout << endl;
}

//二次元配列の表示
void printDimention(DVector &dv){
    rep(i, (int)dv.size()){
        rep(j, (int)dv[i].size()){
            if(j)   cout << " ";
            cout << dv[i][j];
        }
        cout << endl;
    }
}

//numの中に何個2がかかってるか
int factor2(int num){
    int cnt = 0;
    while(num % 2 == 0){
        num /= 2;
        cnt++;
    }
    return cnt;
}