#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <queue>
/*f( i , j ) - > 当前开的箱子的集合为 i , 金钥匙的数目是 j 时可以获得的最多万能钥匙数目.
*/using namespace std;
const int maxn = 15;
int n,sta,stb,stc,f[1 << 15][ maxn*11 ],ans = 0;
typedef struct Item
{int needA,needB,getA,getB,getC;
};typedef struct updatastatus
{int st,number;    updatastatus(const int &st,const int &number){this->st = st , this->number = number;}
};bool arrived[1 << 15][maxn * 11];
queue<updatastatus>q;
Item A[maxn+3];inline void updata(int i,int j,int newans)
{f[i][j] = max(f[i][j],newans);
}//可以开启为 0 ， 不能开启为 1； int main(int argc,char *argv[])
{scanf("%d",&n);for(int i = 0 ; i < n ; ++ i) scanf("%d%d%d%d%d",&A[i].needA,&A[i].needB,&A[i].getA,&A[i].getB,&A[i].getC);scanf("%d%d%d",&sta,&stb,&stc);f[0][sta] = stc; // Init;memset(arrived,false,sizeof(arrived));q.push(updatastatus(0,sta));while(!q.empty()){updatastatus ns = q.front();q.pop();int ra = ns.number, rb, rc = f[ns.st][ns.number] , all = sta + stb + stc , st = ns.st;for(int i = 0 ; i < n ; ++ i)if (ns.st >> i & 1)all += A[i].getA + A[i].getB + A[i].getC - A[i].needA - A[i].needB;ans = max(ans,all);rb = all - ra - rc;for(int i = 0 ; i < n ; ++ i){if (!(st >> i & 1)){int needA = A[i].needA;int needB = A[i].needB;int getA = A[i].getA;int getB = A[i].getB;int getC = A[i].getC;if (ra >= needA) //金够
                        {if (rb >= needB) //金银都够
                           {updata(st | (1 << i) , ra - needA + getA, rc + getC);if (!arrived[st | (1 << i)][ra - needA + getA]){q.push(updatastatus(st | (1 << i),ra - needA + getA));arrived[st | (1 << i)][ra - needA + getA] = true;}    }else  //金够银不够
                  {if (needB- rb <= rc){updata(st | (1 << i) , ra - needA + getA, rc - needB + rb + getC);if (!arrived[st | (1 << i)][ra - needA + getA]){q.push(updatastatus(st | (1 << i),ra - needA + getA));arrived[st | (1 << i)][ra - needA + getA] = true;}    }}}else{if (rb >= needB) //金不够银够
                     {if (needA - ra <= rc){updata(st | (1 << i) , getA, rc - needA + ra + getC);if (!arrived[st | (1 << i)][getA]){q.push(updatastatus(st | (1 << i),getA));arrived[st | (1 << i)][getA] = true;}    }}else  //金不够银不够
                  {if (needA - ra + needB - rb <= rc){updata(st | (1 << i) , getA, rc - needA + ra - needB + rb + getC);if (!arrived[st | (1 << i)][getA]){q.push(updatastatus(st | (1 << i),getA));arrived[st | (1 << i)][getA] = true;}    }      } }}}}printf("%d\n",ans);return 0;
}