// clang-format off #include <bits/stdc++.h> using namespace std; using ll = long long; template<class T> using V = vector<T>; using VI = V<int>; using VL = V<ll>; #define overload4(_1,_2,_3,_4,name,...) name #define rep1(n) for(ll i=0;i<(n);++i) #define rep2(i,n) for(ll i=0;i<(n);++i) #define rep3(i,a,b) for(ll i=(a);i<(b);++i) #define rep4(i,a,b,c) for(ll i=(a);i<(b);i+=(c)) #define rep(...) overload4(__VA_ARGS__,rep4,rep3,rep2,rep1)(__VA_ARGS__) #define all(a) a.begin(),a.end() constexpr ll INF = 1000000000; constexpr ll LLINF = 1LL << 61; template<class T> inline bool chmin(T& a, const T& b) { if (a > b) { a = b; return true; }return false; } inline void init() { cin.tie(nullptr); cout.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(15); } template<class T> inline istream& operator>>(istream& is, V<T>& v) { for (auto& a : v)is >> a; return is; } // clang-format on using pll = pair<ll, ll>; ll cross(pll a, pll b) { return a.first * b.second - a.second * b.first; } int ccw(pll a, pll b, pll c) { b.first -= a.first; b.second -= a.second; c.first -= a.first; c.second -= a.second; if (cross(b, c) == 0) return 0; return cross(b, c) > 0 ? 1 : -1; } int main() { init(); int N, Q; cin >> N >> Q; VL A(N); cin >> A; // Bをソートして累積和を用意する VL B(N - 1); rep(i, N - 1) B[i] = A[i + 1] - A[i]; sort(all(B)); VL C(N); rep(i, N - 1) C[i + 1] = C[i] + B[i]; // 上側凸包を求める V<pll> P; int j = 0; rep(i, N) { while (2 <= j && ccw(P[j - 2], P[j - 1], pll(i, A[i])) >= 0) { --j; P.pop_back(); } P.emplace_back(i, A[i]); ++j; } V<pll> D; // 傾きdy/dxを(dx,dy)のペアとして保持する rep(i, P.size() - 1) { ll dx = P[i + 1].first - P[i].first; ll dy = P[i + 1].second - P[i].second; D.emplace_back(dx, dy); } D.emplace_back(1, -INF); while (Q--) { ll d; cin >> d; int ng = -1, ok = (int)D.size() - 1; while (ok - ng > 1) { int mid = ok + ng >> 1; if (D[mid].second <= d * D[mid].first) ok = mid; // D[mid].second / D[mid].first <= d else ng = mid; } int x = P[ok].first; ll A0 = A[x] - d * (x + 1); ll B0 = A[0] - A0; int up = upper_bound(all(B), d) - B.begin(); cout << (up * d - C[up]) + (d - B0) << "\n"; } return 0; }