human
Write me a Python code to print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k given the following:
PYTHON3 solution for "You are given an integer array a_1, a_2, ..., a_n and integer k.

In one step you can 

  * either choose some index i and decrease a_i by one (make a_i = a_i - 1).

  * or choose two indices i and j and set a_i equal to a_j (make a_i = a_j). 


What is the minimum number of steps you need to make the sum of array ∑_{i=1}^{n}{a_i} ≤ k? (You are allowed to make values of array negative).

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5.
1 ≤ k ≤ 10^{15}) — the size of array a and upper bound on its sum.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array itself.

It's guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k.

Example

Input

4
1 10
20
2 69
6 9
7 8
1 2 1 3 1 2 1
10 1
1 2 3 1 2 6 1 6 8 10

Output

10
0
2
7

Note

In the first test case, you should decrease a_1 10 times to get the sum lower or equal to k = 10.

In the second test case, the sum of array a is already less or equal to 69, so you don't need to change it.

In the third test case, you can, for example: 

  1. set a_4 = a_3 = 1.

  2. decrease a_4 by one, and get a_4 = 0. 

As a result, you'll get array [1, 2, 1, 0, 1, 2, 1] with sum less or equal to 8 in 1 + 1 = 2 steps.

In the fourth test case, you can, for example: 

  1. choose a_7 and decrease in by one 3 times.
you'll get a_7 = -2.

  2. choose 4 elements a_6, a_8, a_9 and a_{10} and them equal to a_7 = -2. 

As a result, you'll get array [1, 2, 3, 1, 2, -2, -2, -2, -2, -2] with sum less or equal to 1 in 3 + 4 = 7 steps.":
assistant
#Here is Python code to print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k.
 def udiv(a,b):
    if a%b==0:
        return a//b
    else:
        return a//b+1
for kfjdkf in range(int(input())):
    n,k=list(map(int,input().split()))
    a=list(map(int,input().split()))
    a.sort()
    sum=0
    for i in a:
        sum+=i
    ans=0
    i=n-1
    mn=a[0]
    if sum<=k:
        print(0)
        continue
    if sum==k+1:
        print(1)
        continue
    if n==1:
        print(max(sum-k,0))
        continue
    b=[]
    s=0
    for i in range(-1,-n-1,-1):
        s+=a[i]
        b.append(s)
    # b.reverse()
    ans=sum-k
    req=sum-k
    # print(ans,req)
    # print(b)
    for i in range(n-1):
        steps=i+1
        dec=b[i]-a[0]*(i+1)
        if dec<req:
            steps+=udiv(req-dec,i+2)
        # print(i,steps)
        # steps=udiv(dec,a[0])
        ans=min(ans,steps)
    print(ans)