#define ATCODER #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; #define FOR(i, a, b) for(ll i=(a); i<(b);i++) #define REP(i, n) for(ll i=0; i<(n);i++) #define ROF(i, a, b) for(ll i=(b-1); i>=(a);i--) #define PER(i, n) for(ll i=n-1; i>=0;i--) #define VL vector #define VVL vector> #define VP vector< pair > #define VVP vector>> #define all(i) begin(i),end(i) #define SORT(i) sort(all(i)) #define EXISTBIT(x,i) (((x>>i) & 1) != 0) #define MP(a,b) make_pair(a,b) #ifdef ATCODER #include using namespace atcoder; using mint = modint1000000007; using mint2 = modint998244353; #endif template vector read(size_t n) { vector ts(n); for (size_t i = 0; i < n; i++) cin >> ts[i]; return ts; } template void read_tuple_impl(TV&) {} template void read_tuple_impl(TV& ts) { get(ts).emplace_back(*(istream_iterator(cin))); read_tuple_impl(ts); } template decltype(auto) read_tuple(size_t n) { tuple...> ts; for (size_t i = 0; i < n; i++) read_tuple_impl(ts); return ts; } template T det2(array ar) { return ar[0] * ar[3] - ar[1] * ar[2]; } template T det3(array ar) { return ar[0] * ar[4] * ar[8] + ar[1] * ar[5] * ar[6] + ar[2] * ar[3] * ar[7] - ar[0] * ar[5] * ar[7] - ar[1] * ar[3] * ar[8] - ar[2] * ar[4] * ar[6]; } template bool chmax(T& tar, T src) { return tar < src ? tar = src, true : false; } template bool chmin(T& tar, T src) { return tar > src ? tar = src, true : false; } template void inc(vector& ar) { for (auto& v : ar) v++; } template void dec(vector& ar) { for (auto& v : ar) v--; } template vector> id_sort(vector& a) { vector res(a.size()); for (int i = 0; i < a.size(); i++)res[i] = MP(a[i], i); SORT(res); return res; } using val = ll; using func = ll; val op(val a, val b) { return a + b; } val e() { return 1; } //val mp(func f, val a) { return MP(a.first + f * a.second, a.second); } //func comp(func f, func g) { return f + g; } //func id() { return 0; } // Rook ll dxr[4] = { 1,0,-1,0 }; ll dyr[4] = { 0,1,0,-1 }; // Bishop ll dxb[4] = { -1,-1,1,1 }; ll djb[4] = { -1,1,-1,1 }; // qween ll dxq[8] = { 0,-1,-1,-1,0,1,1,1 }; ll dyq[8] = { -1,-1,0,1,1,1,0,-1 }; double Power(double var, ll p) { if (p == 0) { return 1.0; } if (p == 1) return var; double ans = Power(var * var, p >> 1); if (p & 1) ans = ans * var;; return ans; } void solve() { // 小数でやってええんなら実験すぎるね ll n, x, y, p, q, rb; cin >> n >> x >> y >> p >> q >> rb; //// i日目で,研修を受けた回数がj回の時の給与最大値 //vector dp(n + 1, vector(n + 1)); //vector rev(n + 1, VL(n + 1, 0)); //PER(i, n) { // REP(j, i + 1) { // // 研修を受ける // if (chmax(dp[i][j], dp[i + 1][j + 1] + p)) { // rev[i][j] = 3; // } // double tmp1 = q; // double tmp2 = q; // double p0 = 1.0 / (x + j * y); // double p1 = 1.0 - p0; // tmp1 += dp[i + 1][j] - r * p0; // tmp2 += dp[i + 1][j] * p1; // // レモンサワーはやっぱりからあげがおすすめですよ // if (chmax(dp[i][j], tmp1)) { // rev[i][j] = 1; // } // // クビになる // if (chmax(dp[i][j], tmp2)) { // rev[i][j] = 2; // } // } //} //REP(i, n) { // REP(j, n) { // cout << rev[i][j] << " "; // } // cout << "\n"; //} // 自明と言えば自明だが // 研修を受ける⇒真面目に作業する⇒クビ上等の三期間に分かれそう // 研修期間:px0 // 真面目期間:x1(q-r(1-p0)) // クビ期間:q(1-p0^x2)/(1-p0) auto calc = [&](ll n0, ll power) { // 真面目期間を三分探索 ll lc = -1, rc = n0; double p0 = 1.0 / power; double p1 = 1.0 - p0; while (lc + 2 < rc) { ll mc0 = (lc * 2 + rc) / 3; ll mc1 = (rc * 2 + lc) / 3; double s0 = mc0 * (q - rb * p0); s0 += p1 >= 1.0 ? q * mc0 : q * (1.0 - Power(p1, (n0 - mc0))) / (1.0 - p1); double s1 = mc1 * (q - rb * p0); s1 += p1 >= 1.0 ? q * mc0 : q * (1.0 - Power(p1, (n0 - mc1))) / (1.0 - p1); if (s0 > s1) { rc = mc1; } else { lc = mc0; } } ll mc = (lc + rc) / 2; double res = mc * (q - rb * p0); res += p1 >= 1.0 ? q * mc : q * (1.0 - Power(p1, (n0 - mc))) / (1.0 - p1);; return res; }; // 研修期間で三分探索 ll l = -1, r = n + 1; while (l + 2 < r) { ll m0 = (l * 2 + r) / 3; ll m1 = (r * 2 + l) / 3; double s0 = calc(n - m0, x + y * m0) + m0 * p; double s1 = calc(n - m1, x + y * m1) + m1 * p; if (s0 > s1) { r = m1; } else { l = m0; } } ll m = (r + l) / 2; double s = calc(n - m, x + y * m) + m * p; printf("%.10f\n", s); return; } int main() { ll t = 1; cin >> t; while (t--) { solve(); } return 0; }