#include using namespace std; #define FOR(i,m,n) for(int i=(m);i<(n);++i) #define REP(i,n) FOR(i,0,n) using ll = long long; constexpr int INF = 0x3f3f3f3f; constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL; constexpr double EPS = 1e-8; constexpr int MOD = 998244353; // constexpr int MOD = 1000000007; constexpr int DY4[]{1, 0, -1, 0}, DX4[]{0, -1, 0, 1}; constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1}; constexpr int DX8[]{0, -1, -1, -1, 0, 1, 1, 1}; template inline bool chmax(T& a, U b) { return a < b ? (a = b, true) : false; } template inline bool chmin(T& a, U b) { return a > b ? (a = b, true) : false; } struct IOSetup { IOSetup() { std::cin.tie(nullptr); std::ios_base::sync_with_stdio(false); std::cout << fixed << setprecision(20); } } iosetup; template struct FenwickTree { explicit FenwickTree(const int n, const Abelian ID = 0) : n(n), ID(ID), data(n, ID) {} void add(int idx, const Abelian val) { for (; idx < n; idx |= idx + 1) { data[idx] += val; } } Abelian sum(int idx) const { Abelian res = ID; for (--idx; idx >= 0; idx = (idx & (idx + 1)) - 1) { res += data[idx]; } return res; } Abelian sum(const int left, const int right) const { return left < right ? sum(right) - sum(left) : ID; } Abelian operator[](const int idx) const { return sum(idx, idx + 1); } int lower_bound(Abelian val) const { if (val <= ID) [[unlikely]] return 0; int res = 0; for (int mask = std::bit_ceil(static_cast(n + 1)) >> 1; mask > 0; mask >>= 1) { const int idx = res + mask - 1; if (idx < n && data[idx] < val) { val -= data[idx]; res += mask; } } return res; } private: const int n; const Abelian ID; std::vector data; }; int main() { int t; cin >> t; while (t--) { int n; cin >> n; FenwickTree bit(n); deque que; ll x = 0; REP(_, n) { int p; cin >> p; --p; if (bit.sum(p) > bit.sum(p + 1, n) || que.empty() || (bit.sum(p) == bit.sum(p + 1, n) && que.front() < p)) { que.emplace_back(p); x += bit.sum(p + 1, n); } else { que.emplace_front(p); x += bit.sum(p); } bit.add(p, 1); } vector a; a.reserve(n); for (const int a_i : que) a.emplace_back(a_i); cout << x << '\n'; REP(i, n) cout << a[i] + 1 << " \n"[i + 1 == n]; } return 0; }