#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;#include <iostream>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <string>
LL exgcd (LL a,LL mod,LL &x,LL &y) {//求解a关于mod的逆元     ★：当且仅当a和mod互质才有用if (mod==0) {x=1;y=0;return a;//保留基本功能，返回最大公约数
    }LL g=exgcd(mod,a%mod,x,y);LL t=x;    //这里根据mod==0  return回来后，x=y;    //x,y是最新的值x2,y2,改变一下，这样赋值就是为了x1=y2y=t-(a/mod)*y;    // y1=x2(变成了t)-[a/mod]y2;return g;            //保留基本功能，返回最大公约数
}
LL get_inv(LL a,LL MOD) //求逆元。记得要a和MOD互质才有逆元的
{LL x,y;  //求a关于MOD的逆元，就是得到的k值是a*k%MOD==1LL GCD=exgcd(a,MOD,x,y);if (GCD==1)//互质才有逆元可说return (x%MOD+MOD)%MOD;//防止是负数else return -1;//不存在
}
const int maxn = 100;
LL r[maxn], mod[maxn];
LL CRT(LL r[], LL mod[], int n) { // X % mod[i] = r[i]LL M = 1;LL ans = 0;for (int i = 1; i <= n; ++i) M *= mod[i];
//    cout << M << endl;for (int i = 1; i <= n; ++i) {LL MI = M / mod[i]; //排除这个数ans += r[i] * (MI * get_inv(MI, mod[i])); //使得MI % mod[i] = 1
//        cout << get_inv(MI, mod[i]) << endl;ans %= M;}
//    if (ans < 0) ans += M;while (true) {int i;for (i = 1; i <= n; ++i) {if (ans < mod[i]) {ans += M;break;}}if (i == n + 1) return ans;}return ans;
}
void work() {int n = 8;for (int i = 1; i <= n; ++i) cin >> mod[i];for (int i = 1; i <= n; ++i) cin >> r[i];cout << CRT(r, mod, n) << endl;
}
int main() {
#ifdef localfreopen("data.txt","r",stdin);
#endifwork();return 0;
}