Codeforces 487A Fight the Monster



A monster is attacking the Cyberland!

Master Yang, a braver, is going to beat the monster. Yang and the monster each have 3 attributes: hitpoints (HP), offensive power (ATK) and defensive power (DEF).

During the battle, every second the monster's HP decrease by max(0, ATK_Y - DEF_M), while Yang's HP decreases bymax(0, ATK_M - DEF_Y), where indexY denotes Master Yang and index M denotes monster. Both decreases happen simultaneously Once monster'sHP ≤ 0 and the same time Master Yang's HP > 0, Master Yang wins.

Master Yang can buy attributes from the magic shop of Cyberland: h bitcoins per HP, a bitcoins per ATK, and d bitcoins per DEF.

Now Master Yang wants to know the minimum number of bitcoins he can spend in order to win.

Input

The first line contains three integers HP_Y, ATK_Y, DEF_Y, separated by a space, denoting the initial HP,ATK and DEF of Master Yang.

The second line contains three integers HP_M, ATK_M, DEF_M, separated by a space, denoting the HP, ATK and DEF of the monster.

The third line contains three integers h, a, d, separated by a space, denoting the price of1 HP, 1 ATK and1 DEF.

All numbers in input are integer and lie between1 and 100 inclusively.

Output

The only output line should contain an integer, denoting the minimum bitcoins Master Yang should spend in order to win.

Examples

Input

1 2 11 100 11 100 100

Output

99

Input

100 100 1001 1 11 1 1

Output

0

Note

For the first sample, prices for ATK andDEF are extremely high. Master Yang can buy99 HP, then he can beat the monster with 1 HP left.

For the second sample, Master Yang is strong enough to beat the monster, so he doesn't need to buy anything.

把所有的攻击力与防御力的可能枚举一遍(所需血量是可以通过这两个值计算出来的)，取最小的花费。

#include<stdio.h>#include<algorithm>using namespace std;struct p{    int atk;    int def;    int cost;}c[20005];bool cmp(p aa,p bb){    return aa.cost<bb.cost;}int main(){    int yh,ya,yd,ma,md,mh,moh,moa,mod;    int s=0;    int cost=0;    scanf("%d%d%d",&yh,&ya,&yd);    scanf("%d%d%d",&mh,&ma,&md);    scanf("%d%d%d",&moh,&moa,&mod);    while (ya<=md)    {        ya++;        cost=cost+moa;    }    for (int i=0;i<=100;i++)    {        for (int j=0;j<=100;j++)        {            s++;            c[s].atk=ya+i;            c[s].def=yd+j;            if ((ma-c[s].def)<=0)            {                c[s].cost=cost+(c[s].def-yd)*mod+(c[s].atk-ya)*moa;                break;            }            if ((mh%(c[s].atk-md))==0)            {                if (yh-(mh/(c[s].atk-md))*(ma-c[s].def)<=0)                {                    c[s].cost=cost+(c[s].def-yd)*mod+(c[s].atk-ya)*moa+moh*(-yh+(mh/(c[s].atk-md))*(ma-c[s].def))+moh;                }                else                {                     c[s].cost=cost+(c[s].def-yd)*mod+(c[s].atk-ya)*moa;                }            }            if ((mh%(c[s].atk-md))!=0)            {                if (yh-(mh/(c[s].atk-md)+1)*(ma-c[s].def)<=0)                {                    c[s].cost=cost+(c[s].def-yd)*mod+(c[s].atk-ya)*moa+moh*(-yh+(mh/(c[s].atk-md)+1)*(ma-c[s].def))+moh;                }                else                {                     c[s].cost=cost+(c[s].def-yd)*mod+(c[s].atk-ya)*moa;                }            }        }    }    sort(c+1,c+s+1,cmp);    printf("%d\n",c[1].cost);}