description: 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.
solution: #include <bits/stdc++.h>
  char ibuf[1 << 15], *p1, *p2;
  struct FastIO {
    inline int rint() {
      register char c, f = 0;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) < 48 ||
             c > 57)
        f |= c == '-';
      register int unsigned a = c & 15;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) >= 48 &&
             c <= 57)
        a = a * 10 + (c & 15);
      return f ? ~a + 1 : a;
    }
    inline long long rll() {
      register char c, f = 0;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) < 48 ||
             c > 57)
        f |= c == '-';
      register unsigned long long a = c & 15;
      while ((c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                                p1 == p2)
                       ? EOF
                       : *p1++)) >= 48 &&
             c <= 57)
        a = a * 10 + (c & 15);
      return f ? ~a + 1 : a;
    }
    inline operator int() { return rint(); }
    inline char rchar() {
      register char c;
      while (!isgraph(
          c = (p1 == p2 && (p2 = (p1 = ibuf) + fread(ibuf, 1, 1 << 15, stdin),
                            p1 == p2)
                   ? EOF
                   : *p1++)))
        ;
      return c;
    }
  } g90;
  std::mt19937 rng(
      std::chrono::high_resolution_clock::now().time_since_epoch().count());
  inline int rnd(const int& a, const int& b) {
    return std::uniform_int_distribution<int>(a, b)(rng);
  }
  inline long double rnd(const long double& a, const long double& b) {
    return std::uniform_real_distribution<long double>(a, b)(rng);
  }
  namespace MY_STD {
  template <typename T>
  inline T abs(const T& a) {
    return a < 0 ? -a : a;
  }
  }  // namespace MY_STD
  const int N = 233333;
  int a[N];
  long long s[N];
  long long divFloor(long long a, long long b) {
    return a < 0 ? (a - b + 1) / b : a / b;
  }
  signed main() {
    for (int T = g90; T--;) {
      int n = g90;
      long long k = g90.rll();
      for (int i = 1; i <= n; ++i) {
        a[i] = g90;
      }
      std::sort(a + 1, a + n + 1);
      for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i];
      if (s[n] <= k) {
        puts("0");
        continue;
      }
      long long ans = 1e18;
      for (int i = 1; i <= n; ++i) {
        ans = std::min(
            ans,
            n - i + std::max(a[1] - divFloor(k - s[i] + a[1], n - i + 1), 0ll));
      }
      printf("%lld\n", ans);
    }
  }
  