这是一道很不错的贪心题目。为了使得开箱的疫苗最少，我们尽量让每个人接种时间靠后，来延长疫苗的过期时间，使得疫苗能够被后面的人接种到。因此对于第一个人 $a_1$，我们就可以规定开箱时间是 $a_1 + w$，那么这箱的过期时间是 $a_1 + w + d$，在这箱过期之前最多能接种 $k$ 个人，如果这之前多于 $k$ 个人，那么只能再开一箱了。这样下去，就能够保证每个人都能接种，并且每箱都能最大化利用，也就是尽量少地开箱。

// Date: Thu Nov 30 21:15:29 2023

#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>

#include <algorithm>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

using namespace std;

const int INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const double eps = 1e-8;
const int dir[8][2] = {
    {0, 1}, {0, -1}, {1, 0}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1},
};

typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> PII;

const ull Pr = 131;

#define For(i, a, b) for (int i = int(a); i < int(b); ++i)
#define Rof(i, a, b) for (int i = int(b) - 1; i >= int(a); --i)
#define For1(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define Rof1(i, a, b) for (int i = int(b); i >= int(a); --i)
#define ForE(i, j) for (int i = h[j]; i != -1; i = ne[i])

#define f1 first
#define f2 second
#define pb push_back
#define has(a, x) (a.find(x) != a.end())
#define nonempty(a) (!a.empty())
#define all(a) (a).begin(), (a).end()
#define SZ(a) int((a).size())

#ifdef _DEBUG
#define debug1(x) cout << #x " = " << x << endl;
#define debug2(x, y) cout << #x " = " << x << " " #y " = " << y << endl;
#define debug3(x, y, z)                                                        \
  cout << #x " = " << x << " " #y " = " << y << " " #z " = " << z << endl;
#else
#define debug1
#define debug2
#define debug3
#endif

const int N = 200010;
int t, n, k, d, w, a[N];

int main(void) {
#ifdef _DEBUG
  freopen("1804b.in", "r", stdin);
#endif
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);

  cin >> t;

  while (t--) {
    cin >> n >> k >> d >> w;

    For1(i, 1, n) { cin >> a[i]; }

    int res = 0, cnt = 0;

    For1(i, 1, n) {
      if (!cnt) {
        int top = a[i] + w + d;
        cnt = 1;
        res++;
        ++i;
        while (cnt < k && i <= n) {
          if (a[i] <= top) {
            cnt++;
            ++i;
          } else
            break;
        }
        --i;
        cnt = 0;
      }
    }

    cout << res << '\n';
  }

  return 0;
}