#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first < b.first) return true;
  if (a.first == b.first) return (a.second < b.second);
  return false;
}
int mex(int *A, int n) {
  int C[n];
  for (int i = 0; i < n; i++) C[i] = A[i];
  sort(C, C + n);
  for (int i = 0; i <= n; i++) {
    if (!binary_search(C, C + n, i)) return i;
  }
}
bool check(int *A, int n) {
  for (int i = 0; i < n - 1; i++) {
    if (A[i] > A[i + 1]) return false;
  }
  return true;
}
bool subsetSum(vector<long long> arr, int n, int s) {
  int dp[2][s + 1];
  for (int i = 0; i <= s; i++) {
    dp[0][i] = false;
  }
  dp[0][0] = true;
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= s; j++) {
      dp[flag][j] = dp[flag ^ 1][j];
      if (j - arr[i - 1] >= 0) {
        dp[flag][j] = dp[flag][j] || dp[flag ^ 1][j - arr[i - 1]];
      }
    }
    flag ^= 1;
  }
  return dp[flag ^ 1][s];
}
void printDFS(int *edges[], int n, int sv, bool *visited) {
  cout << sv << " ";
  visited[sv] = true;
  for (int i = 0; i < n; i++) {
    if (visited[i]) continue;
    if (edges[sv][i] == 1) {
      printDFS(edges, n, i, visited);
    }
  }
}
void DFS(int **edges, int n) {
  bool *visited = new bool[n];
  for (int i = 0; i < n; i++) {
    visited[i] = false;
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) printDFS(edges, n, i, visited);
  }
  delete[] visited;
}
void printBFS(vector<set<int>> edges, int n, int sv, bool *visited) {
  queue<int> q;
  q.push(sv);
  while (!q.empty()) {
    sv = q.front();
    visited[sv] = true;
    q.pop();
    cout << sv << " ";
    for (auto i : edges[sv]) {
      if (!visited[i]) {
        q.push(i);
        visited[i] = true;
      }
    }
  }
  return;
}
void BFS(vector<set<int>> edges, int n) {
  bool *visited = new bool[n];
  for (int i = 0; i < n; i++) {
    visited[i] = false;
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) printBFS(edges, n, i, visited);
  }
  delete[] visited;
}
void buildTree(int *arr, int *tree, int start, int end, int treeNode) {
  if (start == end) {
    tree[treeNode] = arr[start];
    return;
  }
  int mid = (start + end) / 2;
  buildTree(arr, tree, start, mid, 2 * treeNode);
  buildTree(arr, tree, mid + 1, end, 2 * treeNode + 1);
  tree[treeNode] = max(tree[2 * treeNode], tree[2 * treeNode + 1]);
}
void updateTree(int *arr, int *tree, int start, int end, int treeNode, int idx,
                int value) {
  if (start == end) {
    arr[idx] = value;
    tree[treeNode] = value;
    return;
  }
  int mid = (start + end) / 2;
  if (idx > mid) {
    updateTree(arr, tree, mid + 1, end, 2 * treeNode + 1, idx, value);
  } else {
    updateTree(arr, tree, start, mid, 2 * treeNode, idx, value);
  }
  tree[treeNode] = max(tree[2 * treeNode], tree[2 * treeNode + 1]);
}
int query(int *tree, int start, int end, int treeNode, int left, int right) {
  if (start > right || end < left) {
    return INT_MIN;
  }
  if (start >= left && end <= right) {
    return tree[treeNode];
  }
  int mid = (start + end) / 2;
  int ans1 = query(tree, start, mid, 2 * treeNode, left, right);
  int ans2 = query(tree, mid + 1, end, 2 * treeNode + 1, left, right);
  return max(ans1, ans2);
}
long long modExp(long long n, long long p, long long c) {
  if (p == 0) {
    return 1;
  } else {
    long long smallans;
    if (p % 2 == 0) {
      smallans = modExp(n, p / 2, c);
      return (smallans * smallans) % c;
    } else {
      smallans = modExp(n, p - 1, c);
      return ((n % c) * (smallans) % c) % c;
    }
  }
}
int main() {
  int T = 1;
  cin >> T;
  while (T--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    int a;
    for (int i = 0; i < n; i++) {
      cin >> a;
      v[i] = a;
    }
    sort(v.begin(), v.end());
    a = v[0];
    long long ctr = 0;
    long long flag = 1;
    for (int i = 1; i < n; i++) {
      if (v[i] > k) {
        flag = 0;
        break;
      }
      ctr += (k - v[i]) / a;
    }
    if (flag)
      cout << ctr << endl;
    else
      cout << 0 << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

— Hey folks, how do you like this problem?

— That'll do it. 

BThero is a powerful magician. He has got n piles of candies, the i-th pile initially contains a_i candies. BThero can cast a copy-paste spell as follows: 

  1. He chooses two piles (i, j) such that 1 ≤ i, j ≤ n and i ≠ j. 
  2. All candies from pile i are copied into pile j. Formally, the operation a_j := a_j + a_i is performed. 



BThero can cast this spell any number of times he wants to — but unfortunately, if some pile contains strictly more than k candies, he loses his magic power. What is the maximum number of times BThero can cast the spell without losing his power?

Input

The first line contains one integer T (1 ≤ T ≤ 500) — the number of test cases.

Each test case consists of two lines: 

  * the first line contains two integers n and k (2 ≤ n ≤ 1000, 2 ≤ k ≤ 10^4); 
  * the second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ k). 



It is guaranteed that the sum of n over all test cases does not exceed 1000, and the sum of k over all test cases does not exceed 10^4.

Output

For each test case, print one integer — the maximum number of times BThero can cast the spell without losing his magic power.

Example

Input


3
2 2
1 1
3 5
1 2 3
3 7
3 2 2


Output


1
5
4

Note

In the first test case we get either a = [1, 2] or a = [2, 1] after casting the spell for the first time, and it is impossible to cast it again.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first < b.first) return true;
  if (a.first == b.first) return (a.second < b.second);
  return false;
}
int mex(int *A, int n) {
  int C[n];
  for (int i = 0; i < n; i++) C[i] = A[i];
  sort(C, C + n);
  for (int i = 0; i <= n; i++) {
    if (!binary_search(C, C + n, i)) return i;
  }
}
bool check(int *A, int n) {
  for (int i = 0; i < n - 1; i++) {
    if (A[i] > A[i + 1]) return false;
  }
  return true;
}
bool subsetSum(vector<long long> arr, int n, int s) {
  int dp[2][s + 1];
  for (int i = 0; i <= s; i++) {
    dp[0][i] = false;
  }
  dp[0][0] = true;
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= s; j++) {
      dp[flag][j] = dp[flag ^ 1][j];
      if (j - arr[i - 1] >= 0) {
        dp[flag][j] = dp[flag][j] || dp[flag ^ 1][j - arr[i - 1]];
      }
    }
    flag ^= 1;
  }
  return dp[flag ^ 1][s];
}
void printDFS(int *edges[], int n, int sv, bool *visited) {
  cout << sv << " ";
  visited[sv] = true;
  for (int i = 0; i < n; i++) {
    if (visited[i]) continue;
    if (edges[sv][i] == 1) {
      printDFS(edges, n, i, visited);
    }
  }
}
void DFS(int **edges, int n) {
  bool *visited = new bool[n];
  for (int i = 0; i < n; i++) {
    visited[i] = false;
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) printDFS(edges, n, i, visited);
  }
  delete[] visited;
}
void printBFS(vector<set<int>> edges, int n, int sv, bool *visited) {
  queue<int> q;
  q.push(sv);
  while (!q.empty()) {
    sv = q.front();
    visited[sv] = true;
    q.pop();
    cout << sv << " ";
    for (auto i : edges[sv]) {
      if (!visited[i]) {
        q.push(i);
        visited[i] = true;
      }
    }
  }
  return;
}
void BFS(vector<set<int>> edges, int n) {
  bool *visited = new bool[n];
  for (int i = 0; i < n; i++) {
    visited[i] = false;
  }
  for (int i = 0; i < n; i++) {
    if (!visited[i]) printBFS(edges, n, i, visited);
  }
  delete[] visited;
}
void buildTree(int *arr, int *tree, int start, int end, int treeNode) {
  if (start == end) {
    tree[treeNode] = arr[start];
    return;
  }
  int mid = (start + end) / 2;
  buildTree(arr, tree, start, mid, 2 * treeNode);
  buildTree(arr, tree, mid + 1, end, 2 * treeNode + 1);
  tree[treeNode] = max(tree[2 * treeNode], tree[2 * treeNode + 1]);
}
void updateTree(int *arr, int *tree, int start, int end, int treeNode, int idx,
                int value) {
  if (start == end) {
    arr[idx] = value;
    tree[treeNode] = value;
    return;
  }
  int mid = (start + end) / 2;
  if (idx > mid) {
    updateTree(arr, tree, mid + 1, end, 2 * treeNode + 1, idx, value);
  } else {
    updateTree(arr, tree, start, mid, 2 * treeNode, idx, value);
  }
  tree[treeNode] = max(tree[2 * treeNode], tree[2 * treeNode + 1]);
}
int query(int *tree, int start, int end, int treeNode, int left, int right) {
  if (start > right || end < left) {
    return INT_MIN;
  }
  if (start >= left && end <= right) {
    return tree[treeNode];
  }
  int mid = (start + end) / 2;
  int ans1 = query(tree, start, mid, 2 * treeNode, left, right);
  int ans2 = query(tree, mid + 1, end, 2 * treeNode + 1, left, right);
  return max(ans1, ans2);
}
long long modExp(long long n, long long p, long long c) {
  if (p == 0) {
    return 1;
  } else {
    long long smallans;
    if (p % 2 == 0) {
      smallans = modExp(n, p / 2, c);
      return (smallans * smallans) % c;
    } else {
      smallans = modExp(n, p - 1, c);
      return ((n % c) * (smallans) % c) % c;
    }
  }
}
int main() {
  int T = 1;
  cin >> T;
  while (T--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> v(n);
    int a;
    for (int i = 0; i < n; i++) {
      cin >> a;
      v[i] = a;
    }
    sort(v.begin(), v.end());
    a = v[0];
    long long ctr = 0;
    long long flag = 1;
    for (int i = 1; i < n; i++) {
      if (v[i] > k) {
        flag = 0;
        break;
      }
      ctr += (k - v[i]) / a;
    }
    if (flag)
      cout << ctr << endl;
    else
      cout << 0 << endl;
  }
  return 0;
}

```