#define _USE_MATH_DEFINES #include using namespace std; #define FOR(i,m,n) for(int i=(m);i<(n);++i) #define REP(i,n) FOR(i,0,n) #define ALL(v) (v).begin(),(v).end() 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) return 0; int res = 0, exponent = 1; while (exponent <= n) exponent <<= 1; for (int mask = exponent >> 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() { constexpr int N = 200000; int n; cin >> n; vector a(n); REP(i, n) cin >> a[i]; sort(ALL(a)); FenwickTree num(N + 1); REP(i, n) num.add(a[i], 1); FenwickTree bit(N + 1); REP(i, n) bit.add(a[i], a[i]); ll ans = 0; REP(i, n) { ans += (n - 1LL - i) * a[i]; for (int j = a[i]; j <= N; j += a[i]) { const int r = min(j + a[i], N + 1); ans -= bit.sum(j, r) - 1LL * num.sum(j, r) * j; } } cout << ans << '\n'; return 0; }