— Hey folks, how do you like this problem? — That’ll do it.
BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains ai candies. BThero can cast a copy-paste spell as follows:
He chooses two piles (i,j) such that 1≤i,j≤n and i≠j. All candies from pile i are copied into pile j. Formally, the operation aj:=aj+ai is performed. BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?
Input The first line contains one integer T (1≤T≤500) — the number of test cases.
Each test case consists of two lines:
the first line contains two integers n and k (2≤n≤1000, 2≤k≤104); the second line contains n integers a1, a2, …, an (1≤ai≤k). It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 104.
Output For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.
Example Input 3 2 2 1 1 3 5 1 2 3 3 7 3 2 2 Output 1 5 4 Note In the first test case we get either a=[1,2] or a=[2,1] after casting the spell for the first time, and it is impossible to cast it again.
分析：输入一队数列，咋给出最大值k, 操作为a[j]=a[j]+a[i]，魔术师的能力是复制，只要不超过k，可以无限制使用，所以先将数列排序，找出最小的数值，
sort(a+1,a+1+n);将其记录，在按从小到大的顺序，进行复制，
for(i=2;i<=n;i++) { if(a[i]>k) { break; } else { s=k-a[i]; sum+=s/minn; } }次数为（k-a[j]）/a[1]
总代码：
在这里插入代码片