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: //package codeforces.edu120;
  
  import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  //Think through the entire logic before jump into coding!
  //If you are out of ideas, take a guess! It is better than doing nothing!
  //Read both C and D, it is possible that D is easier than C for you!
  
  //Be aware of integer overflow!
  //If you find an answer and want to return immediately, don't forget to flush before return!
  
  public class C {
      static InputReader in;
      static PrintWriter out;
  
      public static void main(String[] args) {
          //initReaderPrinter(true);
          initReaderPrinter(false);
          solve(in.nextInt());
          //solve(1);        
      }
  
      static void solve(int testCnt) {
          for (int testNumber = 0; testNumber < testCnt; testNumber++) {
              int n = in.nextInt();
              long k = in.nextLong();
              Long[] a = in.nextLongArray(n);
              long sum = 0;
              for(int i = 0; i < n; i++) {
                  sum += a[i];
              }
              if(sum <= k) out.println(0);
              else {
                  Arrays.sort(a);
                  long[] ps = new long[n];
                  for(int i = 0; i < n; i++) {
                      ps[i] = (i > 0 ? ps[i - 1] : 0) + a[i];
                  }
                  long ans = Long.MAX_VALUE;
                  for(int setToMin = 1; setToMin <= n; setToMin++) {
                      int nEqualToMin = setToMin - 1;
                      //subbed: the sum difference after replacing the largest nEqualToMin numbers with a[0].
                      long subbed = ps[n - 1] - ps[n - 1 - nEqualToMin] - a[0] * nEqualToMin;
                      //stillNeed: the min sum difference needed to drop the total sum of a to be <= k
                      long stillNeed = max(0, sum - k - subbed);
                      long newMin = a[0] - (stillNeed / setToMin + (stillNeed % setToMin == 0 ? 0 : 1));
                      long ops = nEqualToMin + a[0] - newMin;
                      ans = min(ans, ops);
                  }
                  out.println(ans);
              }
          }
          out.close();
      }
  
      static void initReaderPrinter(boolean test) {
          if (test) {
              try {
                  in = new InputReader(new FileInputStream("src/input.in"));
                  out = new PrintWriter(new FileOutputStream("src/output.out"));
              } catch (IOException e) {
                  e.printStackTrace();
              }
          } else {
              in = new InputReader(System.in);
              out = new PrintWriter(System.out);
          }
      }
  
      static class InputReader {
          BufferedReader br;
          StringTokenizer st;
  
          InputReader(InputStream stream) {
              try {
                  br = new BufferedReader(new InputStreamReader(stream), 32768);
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          Integer[] nextIntArray(int n) {
              Integer[] a = new Integer[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitive(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          int[] nextIntArrayPrimitiveOneIndexed(int n) {
              int[] a = new int[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextInt();
              return a;
          }
  
          Long[] nextLongArray(int n) {
              Long[] a = new Long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitive(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long[] nextLongArrayPrimitiveOneIndexed(int n) {
              long[] a = new long[n + 1];
              for (int i = 1; i <= n; i++) a[i] = nextLong();
              return a;
          }
  
          String[] nextStringArray(int n) {
              String[] g = new String[n];
              for (int i = 0; i < n; i++) g[i] = next();
              return g;
          }
  
          List<Integer>[] readGraphOneIndexed(int n, int m) {
              List<Integer>[] adj = new List[n + 1];
              for (int i = 0; i <= n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt();
                  int v = nextInt();
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          List<Integer>[] readGraphZeroIndexed(int n, int m) {
              List<Integer>[] adj = new List[n];
              for (int i = 0; i < n; i++) {
                  adj[i] = new ArrayList<>();
              }
              for (int i = 0; i < m; i++) {
                  int u = nextInt() - 1;
                  int v = nextInt() - 1;
                  adj[u].add(v);
                  adj[v].add(u);
              }
              return adj;
          }
  
          /*
              A more efficient way of building a graph using int[] instead of ArrayList to store each node's neighboring nodes.
              1-indexed.
           */
          int[][] buildGraph(int nodeCnt, int edgeCnt) {
              int[] end1 = new int[edgeCnt], end2 = new int[edgeCnt];
              int[] edgeCntForEachNode = new int[nodeCnt + 1], idxForEachNode = new int[nodeCnt + 1];
              for (int i = 0; i < edgeCnt; i++) {
                  int u = in.nextInt(), v = in.nextInt();
                  edgeCntForEachNode[u]++;
                  edgeCntForEachNode[v]++;
                  end1[i] = u;
                  end2[i] = v;
              }
              int[][] adj = new int[nodeCnt + 1][];
              for (int i = 1; i <= nodeCnt; i++) {
                  adj[i] = new int[edgeCntForEachNode[i]];
              }
              for (int i = 0; i < edgeCnt; i++) {
                  adj[end1[i]][idxForEachNode[end1[i]]] = end2[i];
                  idxForEachNode[end1[i]]++;
                  adj[end2[i]][idxForEachNode[end2[i]]] = end1[i];
                  idxForEachNode[end2[i]]++;
              }
              return adj;
          }
      }
  }