Ferries Wheel

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)


The Ferries Wheel is a circle,rounded by many cable cars,and the cars are numbered $1,2,3...K-1,K$ in order.Every cable car has a unique value and $A[i-1] < A[i] < A[i+1] (1<i<K)$.

Today,Misaki invites $N$ friends to play in the Ferries Wheel.Every one will enter a cable car. One person will receive a kiss from Misaki,if this person satisfies the following condition: (his/her cable car's value + the left car's value
) % INT_MAX = the right car's value,the ${1}^{st}$ car’s left car is the ${k}^{th}$ car,and the right one is ${2}^{nd}$ car,the ${k}^{th}$ car’s left car is the ${(k-1)}^{th}$ car,and the right one is the ${1}^{st}$ car.

Please help Misaki to calculate how many kisses she will pay,you can assume that there is no empty cable car when all friends enter their cable cars,and one car has more than one friends is valid.


There are many test cases.
For each case,the first line is a integer $N(1<=N<=100)$ means Misaki has invited $N$ friends,and the second line contains $N$ integers $val1,val2,...valN$, the $val[i]$ means the ${i}^{th}$ friend's cable car's value.
$(0<= val[i] <=$ INT_MAX).

The INT_MAX is $2147483647$.


For each test case, first output Case #X: ,then output the answer, if there are only one cable car, print "-1";

Sample Input

3 1 2 3 5 1 2 3 5 7 6 2 3 1 2 7 5

Sample Output

Case #1: 1 Case #2: 2 Case #3: 3
In the third sample, the order of cable cars is {{1},{2}, {3}, {5}, {7}} after they enter cable car,but the 2nd cable car has 2 friends,so the answer is 3.




Valentine's Day Round