func_code_string: #include <bits/stdc++.h>   using namespace std;   inline long long max(long long a, long long b) { return ((a > b) ? a : b); }   inline long long min(long long a, long long b) { return ((a > b) ? b : a); }   inline long long gcd(long long a, long long b) {     if (b == 0) return a;     a %= b;     return gcd(b, a);   }   long long power(long long x, long long y) {     if (y == 0) return 1;     long long a = power(x, y / 2);     if (y % 2 == 0)       return a * a;     else       return x * a * a;   }   long long powermod(long long x, long long ex, long long md) {     long long ans = 1ll;     while (ex > 0) {       if (ex & 1ll) ans = (ans * x) % md;       ex >>= 1ll;       x = (x * x) % md;     }     return ans;   }   const long long inf = 1e18 + 9;   const long long mod = 1e9 + 7;   const long double PI = acos(-1);   const long double eps = 1e-9;   const long long N = 2e5 + 11;   long long n, t, easy, hard;   long long a[N], b[N], pfx[N], cnteasy[N], cnthard[N];   vector<pair<long long, long long> > v;   void solve() {     cin >> n >> t >> easy >> hard;     for (int i = 0; i < n; i++) cin >> b[i + 1];     for (int i = 0; i < n; i++) cin >> a[i + 1];     for (int i = 0; i < n; i++) v.push_back(make_pair(a[i + 1], b[i + 1]));     sort((v).begin(), (v).end());     for (int i = 0; i < n; i++) {       if (v[i].second)         pfx[i + 1] = pfx[i] + hard;       else         pfx[i + 1] = pfx[i] + easy;       cnthard[i + 1] = cnthard[i] + (v[i].second);       cnteasy[i + 1] = cnteasy[i] + (!v[i].second);     }     if (pfx[n] <= t) {       cout << n <<   n ;     } else {       long long ans = 0;       for (long long i = 0; i < n; i++) {         long long t = v[i].first - 1;         if (t < 0) continue;         long long val = i;         long long tym = pfx[i];         long long rem = t - tym;         if (rem >= 0) {           long long x = min(rem / easy, cnteasy[n] - cnteasy[i]);           rem -= (x * easy);           val += x;           val += min(rem / hard, cnthard[n] - cnthard[i]);           ans = max(ans, val);         }       }       cout << ans <<   n ;     }     for (int i = 0; i < n + 1; i++)       a[i] = b[i] = pfx[i] = cnteasy[i] = cnthard[i] = 0;     v.clear();     n = t = easy = hard = 0;   }   int main() {     {       ios::sync_with_stdio(false);       cin.tie(0);       cout.tie(0);     };     cout << fixed << setprecision(20);     long long NTC = 1;     cin >> NTC;     long long PTC = 0;     while (PTC++ < NTC) {       solve();       cout <<   n ;     }     cerr <<  Time :   << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC          <<  ms n ;   }   