#include #include using namespace std; using namespace atcoder; template struct StaticModint { using mint = StaticModint; public: int val; StaticModint() : val(0) {} StaticModint(long long v) { if (std::abs(v) >= mod()) { v %= mod(); } if (v < 0) { v += mod(); } val = v; } mint &operator++() { val++; if (val == mod()) { val = 0; } return *this; } mint &operator--() { if (val == 0) { val = mod(); } val--; return *this; } mint &operator+=(const mint &x) { val += x.val; if (val >= mod()) { val -= mod(); } return *this; } mint &operator-=(const mint &x) { val -= x.val; if (val < 0) { val += mod(); } return *this; } mint &operator*=(const mint &x) { val = (int)((long long)val * x.val % mod()); return *this; } mint &operator/=(const mint &x) { *this *= x.inv(); return *this; } mint operator-() { return mint() - *this; } mint pow(long long n) const { mint x = 1, r = *this; while (n) { if (n & 1) { x *= r; } r *= r; n >>= 1; } return x; } mint inv() const { return pow(mod() - 2); } friend mint operator+(const mint &x, const mint &y) { return mint(x) += y; } friend mint operator-(const mint &x, const mint &y) { return mint(x) -= y; } friend mint operator*(const mint &x, const mint &y) { return mint(x) *= y; } friend mint operator/(const mint &x, const mint &y) { return mint(x) /= y; } friend bool operator==(const mint &x, const mint &y) { return x.val == y.val; } friend bool operator!=(const mint &x, const mint &y) { return x.val != y.val; } friend std::ostream &operator<<(std::ostream &os, const mint &x) { return os << x.val; } friend std::istream &operator>>(std::istream &is, mint &x) { long long v; is >> v; x = mint(v); return is; } private: static constexpr int mod() { return MOD; } }; using mint = StaticModint<998244353>; int main() { int n; cin >> n; int p[200005], q[200005]; for(int i = 0; i < n; i++) { cin >> p[i]; p[i]--; q[p[i]] = i; } mint ans = 0; fenwick_tree bit0(n), bit1(n); for(int i = 0; i < n; i++) { ans += bit0.sum(q[i], n) - bit1.sum(q[i], n) * mint(2).pow(q[i]); bit0.add(q[i], mint(2).pow(n - 1)); bit1.add(q[i], mint(2).pow(n - 1 - q[i])); } cout << ans << endl; }