#include // {{{ // https://codeforces.com/blog/entry/96344 #pragma GCC optimize("O2,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") using namespace std; template [[nodiscard]] static T MIN() { return std::numeric_limits::min(); } template [[nodiscard]] static T MAX() { return std::numeric_limits::max(); } template [[nodiscard]] static T sc(auto&& x) { return static_cast(x); } template [[nodiscard]] static T sz(auto&& x) { return static_cast(x.size()); } #define prln(...) std::println(__VA_ARGS__) #define pr(...) std::print(__VA_ARGS__) #ifdef LOCAL #define dbgln(...) std::println(__VA_ARGS__) #define dbg(...) std::print(__VA_ARGS__) #endif inline static void NO() { prln("NO"); } inline static void YES() { prln("YES"); } using ll = long long; using ld = long double; template using ve = std::vector; template using ar = std::array; template using pa = std::pair; template using tu = std::tuple; template using dq = std::deque; template using qu = std::queue; template using pq = std::priority_queue; template using st = std::stack; #define ff first #define ss second #define eb emplace_back #define pb push_back #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() // }}} void solve() { int n; cin >> n; ve> a(n); for (int i = 0; i < n; ++i) { cin >> a[i].ff; a[i].ss = i; } sort(all(a)); ve prefix(n + 1, 0), postfix(n + 2, 0); for (int i = 0; i < n; ++i) { prefix[i + 1] = prefix[i] + a[i].ff; } for (int i = n - 1; i >= 0; --i) { postfix[i + 1] = postfix[i + 2] + a[i].ff; } ve ans(n); for (int i = 0; i < n; ++i) { ll x = n + a[i].ff * (i + 1) - prefix[i + 1]; x += postfix[i + 2] - a[i].ff * (n - i - 1); ans[a[i].ss] = x; } for (auto& e : ans) pr("{} ", e); prln(); /* 1 3 4 3 + 0 + to query @ i, want to get info related to a[i] - a[0], a[i] - a[1], .... prefix differences + postfix differences for each index n + a[i] - a[0] + a[i] - a[1] + ... + a[i] - a[i] + a[j] - a[i] + a[j + 1] - a[i] = n + a[i] * (i + 1) - prefix[i] (inclusive) + postfix[i] (exclusive) - a[i] * `` n * a[i] - sum of prefix before i + sum of postfix after i sum a[i] - a[j] + 1 for j in range(i + 1) count how manny points before us? * a[i] remove prefix sum up to j - and postfix! <- note this *-** - --- ---- --- - -- ---- -- - really, we're just adding the inclusive distance between ALL pairs */ } int main() { // {{{ cin.tie(nullptr)->sync_with_stdio(false); cin.exceptions(cin.failbit); int t = 1; cin >> t; while (t--) { solve(); } return 0; } // }}}