#include <stdio.h>
/*** @author 杨文蓁的小迷弟* @note   算法作业3* @date   2019/09/05*/
int main()
{//北大OJ数据超过了正常的int，估计有一个50000的数据unsigned int goods1, goods2, goods3, goods4, goods5, goods6; //1，2，3，4，5，6while (scanf("%d %d %d %d %d %d", &goods1, &goods2, &goods3, &goods4, &goods5, &goods6) != EOF){unsigned int sum = 0, spaceby1 = 0, spaceby2 = 0; //sum总箱子数，n余下放1x1的位置，m余下放2x2的位置if (goods1 == 0 && goods2 == 0 && goods3 == 0 && goods4 == 0 && goods5 == 0 && goods6 == 0){return 0;}sum += goods4 + goods5 + goods6 + ((goods3 +3) / 4); //最少需要开的箱子if (goods3 % 4 == 1){spaceby1 += 7;spaceby2 += 5;} //余下1个3*3所余下的1*1和2*2的空间if (goods3 % 4 == 2){spaceby1 += 6;spaceby2 += 3;} //余下2个3*3所余下的1*1和2*2的空间if (goods3 % 4 == 3){spaceby1 += 5;spaceby2 += 1;}           //余下3个3*3所余下的1*1和2*2的空间spaceby2 += 5 * goods4; //装一个4*4的时候余下的1*1和2*2的空间if (spaceby2 <= goods2) //2*2的空间小于2*2的盒子
        {//额外的2*2空间需要的箱子sum += ((goods2 - spaceby2 + 8) / 9);}//余下了多少 1*1 的空间spaceby1 = 36 * sum - 36*goods6 - 25*goods5 - 16*goods4 - 9*goods3 - 4*goods2;if (spaceby1 <= goods1){sum += ((goods1 - spaceby1 + 35) / 36);}printf("%d\n", sum);}return 0;
}

#author: 杨文蓁的小迷弟
#date:   2019/9/5while (1):s=input().split()box_sum = spaceby1 = spaceby2 = 0flag = Truefor i in range(len(s)):if int(s[i]) != 0:flag = Falseif flag == True:exit()box_sum = int(s[5]) + int(s[4]) + int(s[3]) + (int(s[2])-1) // 4+1if int(s[2])%4 == 1:spaceby1 += 5spaceby2 += 7if int(s[2])%4 == 2:spaceby1 += 6spaceby2 += 3if int(s[2])%4 == 3:spaceby1 += 5spaceby2 += 1spaceby2 += 5 * int(s[3])if spaceby2 <= int(s[1]):box_sum += ((int(s[1])-spaceby2-1)//9 + 1)spaceby1 = 36*box_sum - 36*int(s[5]) - 25*int(s[4]) - 16*int(s[3]) - 9*int(s[2]) - 4*int(s[1]);if spaceby1 <= int(s[0]):box_sum += ((int(s[0])-spaceby1-1)//36 + 1)print(box_sum);