# Marks Distribution

In an examination one student appeared in N subjects and has got total T marks. He has passed in all the N subjects where minimum mark for passing in each subject is P. You have to calculate the number of ways the student can get the marks. For example, if N = 3, T = 34 and P = 10 then the marks in the three subject could be as follows. Subject 1 114 213 313 412 512 611 711 810 910 10 11 11 10 12 12 13 10 14 11 15 10 Subject 2 10 11 10 11 10 11 10 11 10 12 12 12 13 13 14 Subject 3 10 10 11 11 12 12 13 13 14 11 12 10 11 10 10 So there are 15 solutions. So F (3, 34, 10) = 15. Input In the first line of the input there will be a single positive integer K followed by K lines each containing a single test case. Each test case contains three positive integers denoting N, T and P respectively. ThevaluesofN,T andP willbe1≤N ≤70,1≤P ≤T ≤70. Youmayassumethatthefinalanswer will fit in a standard 32-bit integer. Output For each input, print in a line the value of F(N,T,P). Sample Input 2 3 34 10 3 34 10 Sample Output 15 15