#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);
  }
}

### Prompt

In CPP, your task is to solve the following problem:

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.

### Response

```cpp
#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);
  }
}

```