The Royal Canadian Mint has
commissioned a new series of designer coffee tables, with legs that are
constructed from stacks of coins. Each table has four legs, each of which uses a
different type of coin. For example, one leg might be a stack of quarters,
another nickels, another loonies, and another twonies. Each leg must be exactly
the same length.
Many coins are available for these tables, including
foreign and special commemorative coins. Given an inventory of available coins
and a desired table height, compute the lengths nearest to the desired height
for which four legs of equal length may be constructed using a different coin
for each leg.

Input consists of several test
cases. Each case begins with two integers: 4 <= n <= 50 giving the number
of types of coins available, and 1 <= t <= 10 giving the number of tables
to be designed. n lines follow; each gives the thickness of a coin in hundredths
of millimetres. t lines follow; each gives the height of a table to be designed
(also in hundredths of millimetres). A line containing 0 0 follows the last test
case.

For each table, output a line
with two integers: the greatest leg length not exceeding the desired length, and
the smallest leg length not less than the desired length.

#include<stdio.h>

#include<algorithm>

using namespace std;

const int N=3e6;

int f[N],a[];

int gcd(int a,int b)

{

    return b?gcd(b,a%b):a;

}

int main()

{

    int n,t;

    while(scanf("%d%d",&n,&t),n||t)

    {

        for(int i=; i<n; i++)

            scanf("%d",&a[i]);

        int b=;

        for(int i=; i<n-; i++)

            for(int j=i+; j<n-; j++)

            {

                int m=a[i]/gcd(a[i],a[j])*a[j];

                for(int k=j+; k<n-; k++)

                {

                    int mm=m/gcd(m,a[k])*a[k];

                    for(int l=k+; l<n; l++)

                        f[b++]=mm/gcd(mm,a[l])*a[l];

                }

            }

        while(t--)

        {

            int q;

            scanf("%d",&q);

            int ma=,mi=0x3f3f3f3f;

            for(int i=; i<b; i++)

                if(q%f[i]==)

                {

                    ma=mi=q;

                    break;

                }

                else

                {

                    int x=q/f[i]*f[i];

                    int y=(q/f[i]+)*f[i];

                    ma=max(ma,x);

                    mi=min(mi,y);

                }

           printf("%d %d\n",ma,mi);

        }

    }

    return ;

}