human
Can you solve the below in PYTHON3?
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.

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What is a Python program to print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k?
assistant
#Here is Python program to print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k.

import sys, collections, math, bisect, heapq, random, functools,io,os,copy
input = sys.stdin.readline
out = sys.stdout.flush
def rep():
    a = list(map(int,input().split()))
    return a

def sep():
    a = list(input().split())
    return a

class UnionFind:
    def __init__(self, x) -> None:
        self.uf = [-1] * x

    def find(self, x):
        r = x
        while self.uf[x] >= 0:
            x = self.uf[x]

        while r != x:
            self.uf[r], r = x, self.uf[r]
        return x

    def union(self, x, y):
        ux, uy = self.find(x), self.find(y)
        if ux == uy:
            return
        if self.uf[ux] >= self.uf[uy]:
            self.uf[uy] += self.uf[ux]
            self.uf[ux] = uy
        else:
            self.uf[ux] += self.uf[uy]
            self.uf[uy] = ux
        return

    def __print__(self):
        return self.uf

def spfa(x,g,n):
    dis = [float('inf') for i in range(n)]
    dis[x] = 0
    state = [False for i in range(n)]
    state[x] = True
    queue = collections.deque()
    queue.append(x)
    while queue:
        cur = queue.popleft()
        state[cur] = False
        for next_ in g[cur]:
            if dis[next_] > dis[cur] + 1:
                dis[next_] = dis[cur] + 1
                if state[next_] == False:
                    state[next_] = True
                    queue.append(next_)
    return dis

def gcd(x,y):
    if y == 0:
        return x
    return gcd(y,x % y)

def valid(x,a,v,total,k):
    # 枚举减少几次
    minv = v[0]
    n = len(v)
    if x > n - 1:
       minv = minv - (x - n + 1)
       total = total - (x - n + 1)
       x = n - 1

    for i in range(x + 1):
        sufsum = a[x - i]
        cursum = total - i - sufsum + (x - i) * (minv - i)
        if cursum <= k:
            return True

    return False
def solve():
    n,k = map(int,input().split())
    a = rep()
    t = sum(a)
    a.sort()
    suf = [0]
    for i in range(n - 1,-1,-1):
        suf.append(suf[-1] + a[i])

    if t <= k:
        print(0)
    else:

        l,r = 1,t - k
  ...