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Many years ago Berland was a small country where only n people lived. Each person had some savings: the i-th one had ai burles.

The government considered a person as wealthy if he had at least x burles. To increase the number of wealthy people Berland decided to carry out several reforms. Each reform looked like that:

the government chooses some subset of people (maybe all of them);

the government takes all savings from the chosen people and redistributes the savings among the chosen people equally. For example, consider the savings as list [5,1,2,1]: if the government chose the 1-st and the 3-rd persons then it, at first, will take all 5+2=7 burles and after that will return 3.5 burles to the chosen people. As a result, the savings will become [3.5,1,3.5,1].

A lot of data was lost from that time, so we don’t know how many reforms were implemented and to whom. All we can do is ask you to calculate the maximum possible number of wealthy people after several (maybe zero) reforms.

Input

The first line contains single integer T (1≤T≤1000) — the number of test cases.

Next 2T lines contain the test cases — two lines per test case. The first line contains two integers n and x (1≤n≤105, 1≤x≤109) — the number of people and the minimum amount of money to be considered as wealthy.

The second line contains n integers a1,a2,…,an (1≤ai≤109) — the initial savings of each person.

It’s guaranteed that the total sum of n doesn’t exceed 105.

Output

Print T integers — one per test case. For each test case print the maximum possible number of wealthy people after several (maybe zero) reforms.

Example

inputCopy 4 4 3 5 1 2 1 4 10 11 9 11 9 2 5 4 3 3 7 9 4 9 outputCopy 2 4 0 3

Note

The first test case is described in the statement.

In the second test case, the government, for example, could carry out two reforms: [11–––,9–,11,9]→[10,10,11–––,9–]→[10,10,10,10].

In the third test case, the government couldn’t make even one person wealthy.

In the fourth test case, the government could choose all people to carry out a reform: [9–,4–,9–]→[713,713,713].

题意

有n个人，每个人都有一定数量的钱，选出一部分人来将他们的钱平分，问在评分后的钱大于x的情况下，最多能选多少人

思路

优先选钱多的，

本来就大于x的一定选，然后再优先选取钱较多的才能实现最大 具体实现见代码

代码：

#include
   
    #include
    
     #include
     
      using namespace std;typedef long long ll;const int N = 1e5 + 10;int t;ll a[N];bool cmp(int x, int y){
       return x > y;}int main() {
       cin>>t;    while (t--)     {
           ll n, x;        cin>>n>>x;        for (int i = 1; i <= n; ++i)            cin>>a[i];        sort(a + 1, a + n + 1,cmp);//降序        ll sum = 0;        int ans = 0;        for (int i = 1; i <= n; ++i) //判断选取        {
               sum += a[i];            if (sum < x * i)                 break;            else                 ans++;        }        printf("%d\n", ans);    }}
     
    
   

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