You are in a maze; seeing n doors in front of you in beginning. You can choose any door you like. The probability for choosing a door is equal for all doors.

If you choose the ith door, it can either take you back to the same position where you begun in xi minutes, or can take you out of the maze after xi minutes. If you come back to the same position, you can't remember anything. So, every time you come to the beginning position, you have no past experience.

Now you want to find the expected time to get out of the maze.
 

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a blank line and an integer n (1 ≤ n ≤ 100) denoting the number of doors. The next line contains n space separated integers. If the ith integer (xi) is positive, you can assume that the ith door will take you out of maze after xi minutes. If it's negative, then the ith door will take you back to the beginning position after abs(xi) minutes. You can safely assume that 1 ≤ abs(xi) ≤ 10000.
 

Output

For each case, print the case number and the expected time to get out of the maze. If it's impossible to get out of the maze, print 'inf'. Print the result in p/q format. Where p is the numerator of the result and q is the denominator of the result and they are relatively prime. See the samples for details.
 

Sample Input

3



1

1



2

-10 -3



3

3 -6 -9

Sample Output

Case 1: 1/1

Case 2: inf

Case 3: 18/1

題目大意：
大概就是每次選擇一個門，如果是正數，則花費ai時間就逃離，如果是負數，則花費-a[i]回到原地，求花費時間的期望

思路

感覺期望的題都是把和的期望換成期望的和，那麼顯然這題可以列出等式
E(x) = E[x1+x2+x3+x4…..+xn) 其中 xi就等於，在第i個門所花費的時間期望，
繼而等於 E(x1) + E(x2) + E(x3)+E(x4)+…….+E(xn)
那麼考慮每一個門的期望，當門是正數時，就是1/n*a[i]
當門是負數時期望是1/n*(a[i]+x) 其中x為期望 ，如果是無限種可能的大概都可以這麼搞
然後化簡就行了

accode

#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define lson rt<<1
#define rson rt<<1|1
using namespace std;
const int maxn = 1e5+53;
const LL mod = 998244353;
int T;
int n;
LL a[maxn];
int main()
{
    scanf("%d",&T);
    int ka = 1;
    while(T--){
        scanf("%d",&n);
        LL sum = 0;
        LL cnt = 0;
        for(int i = 1;i<=n;i++){
            scanf("%lld",&a[i]);
            sum+=abs(a[i]);
            if(a[i]>=0){
                cnt++;
            }
        }
        printf("Case %d: ",ka++);
        if(cnt<=0){
            puts("inf");
            continue;
        }
         LL gcd = __gcd(sum,cnt);
         printf("%lld/%lld\n",sum/gcd,cnt/gcd);
    }
}