In ‘MonkeyLand‘, there is a traditional game called "Bamboo Climbing". The rules of the game are as follows:

1)       There are N monkeys who play this game and there are N bamboos of equal heights. Let the height be L meters.

2)       Each monkey stands in front of a bamboo and every monkey is assigned a different bamboo.

3)       When the whistle is blown, the monkeys start climbing the bamboos and they are not allowed to jump to a different bamboo throughout the game.

4)       Since they are monkeys, they usually climb by jumping. And in each jump, the i^th monkey can jump exactly p_i meters (p_i is a prime). After a while when a monkey finds
that he cannot jump because one more jump may get him out of the bamboo, he reports the remaining length r_i that he is not able to cover.

5)       And before the game, each monkey is assigned a distinct p_i.

6)       The monkey, who has the lowest r_i, wins.

Now, the organizers have found all the information of the game last year, but unluckily they haven‘t found the height of the bamboo. To be more exact, they knowN, all p_i and corresponding r_i,
but not L. So, you came forward and found the task challenging and so, you want to find L, from the given information.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 12). Each of the next n lines contains two integers p_i (1 < p_i < 40, p_i is a prime) and r_i(0
< r_i < p_i). All p_i will be distinct.

Output

For each case, print the case number and the minimum possible value of L that satisfies the above conditions. If there is no solution, print ‘Impossible‘.

题意：告诉你猴子爬杆子，每次跳素数长度的距离p[i]，最后剩余r[i],问你杆子长度

分析：首先我们先学习下中国剩余定理：中国剩余定理

在模数mi两两互素的情况下，可以用中国剩余定理求n使其满足：

n=b1(mod m1)

n=b2(mod m2)

n=b3(mod m3)

....

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<set>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
const int INF=0x3f3f3f3f;
typedef long long LL;
const int maxn=1e6+100;
void exgcd(LL a,LL b,LL &xx,LL &yy)
{
    if(b==0)
    {
        xx=1;yy=0;
        return;
    }
    exgcd(b,a%b,xx,yy);
    LL t=xx;xx=yy;yy=t-a/b*yy;
    return;

}
LL China(LL m[],LL b[],int k)//m:模数 b:余数
{
    LL n=1,xx,yy;
    LL ans=0;
    for(int i=0;i<k;i++)
        n*=m[i];
    for(int i=0;i<k;i++)
    {
        LL t=n/m[i];
        exgcd(t,m[i],xx,yy);
        ans=(ans+xx*t*b[i])%n;
    }
    return (ans%n+n)%n;
}
LL p[45],r[45];
int main()
{
    int t,n,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        REP(i,n)
            scanf("%lld%lld",&p[i],&r[i]);
        LL ans=China(p,r,n);
        printf("Case %d: %lld\n",cas++,ans);
    }
    return 0;
}

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