HDU 1788: Chinese remainder theorem again

#include <iostream>
#include <string>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define LL long long
#define CLS(arr) memset(arr,0,sizeof(arr))
using namespace std;
LL l,n;
LL gcd(LL a,LL b){
    return !b?a:gcd(b,a%b);
}
void solve(){
    while(cin>>l>>n,l,n){
        LL lcm=1,num;
        range(i,1,l){
            cin>>num;
            lcm=lcm*num/gcd(lcm,num);
        }
        cout<<lcm-n<<endl;
    }
}
int main(int argc, char *argv[]){
    solve();
    return 0;
}