#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("O3")
#pragma GCC optimize("O2")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (long long i = 0; i < v.size(); ++i) os << v[i] << " ";
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v) {
  for (auto it : v) os << it << " ";
  return os;
}
template <typename T, typename S>
ostream& operator<<(ostream& os, const pair<T, S>& v) {
  os << v.first << " " << v.second;
  return os;
}
const long long mod = 1e9 + 7;
const long long inf = 2e18;
const long long ninf = -2e18;
long long pow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  time_t t1, t2;
  t1 = clock();
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) {
    long long n, x, y, d;
    cin >> n >> x >> y >> d;
    long long can = false;
    long long ans = -1;
    if (x > y) {
      if ((x - y) % d == 0) {
        can = true;
        ans = (x - y) / d;
      }
    } else {
      if ((y - x) % d == 0) {
        can = true;
        ans = (y - x) / d;
      }
    }
    if (!can) {
      if ((n - y) % d == 0) {
        can = true;
        if ((n - x) % d == 0) {
          ans = (n - x) / d;
        } else
          ans = (n - x) / d + 1;
        ans += (n - y) / d;
      }
      if ((y - 1) % d == 0) {
        long long tans;
        if ((x - 1) % d == 0) {
          tans = (x - 1) / d;
        } else {
          tans = (x - 1) / d + 1;
        }
        tans += (y - 1) / d;
        if (can) {
          if (tans < ans) ans = tans;
        } else
          ans = tans;
      }
    }
    cout << ans << endl;
  }
  t2 = clock();
  cerr << endl << t2 - t1 << endl;
  return 0;
}
