human
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 better_floor(a, b):
    c = a // b
    while (c * b) > a:
        c -= 1
    return c

def solve(a, n, k):
    result = 1e18
    a.sort()
    # Create prefix sums
    pSum = [a[0]]
    for ii in range(1, len(a)):
        pSum.append(pSum[ii-1]+a[ii])

    # calculate minimum steps for each y value in range(n)
    for y in range(len(a)):
        # print("testing y:", y)
        x = a[0] - better_floor(k - pSum[n-1-y] + a[0], y+1)   # This is the equation I was missing
        # x = a[0] - ( (k - pSum[n-1-y] + a[0]) // (y+1) )   # This is the equation I was missing
        # print("with x", x, "and y", y, "we have moves:", x+y)
        result = min(result, max(0, x) + y)
        # if result < 0:
        #     print(0)
        #     return

    print(result)

for test in range(int(input())):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))

    solve(a, n, k)

"""
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

"""