#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; constexpr int INF = 1001001001; constexpr int mod = 1000000007; // constexpr int mod = 998244353; template inline bool chmax(T& x, T y){ if(x < y){ x = y; return true; } return false; } template inline bool chmin(T& x, T y){ if(x > y){ x = y; return true; } return false; } struct mint { int x; mint() : x(0) {} mint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {} mint& operator+=(const mint& p){ if((x += p.x) >= mod) x -= mod; return *this; } mint& operator-=(const mint& p){ if((x -= p.x) < 0) x += mod; return *this; } mint& operator*=(const mint& p){ x = (int)(1LL * x * p.x % mod); return *this; } mint& operator/=(const mint& p){ *this *= p.inverse(); return *this; } mint operator-() const { return mint(-x); } mint operator+(const mint& p) const { return mint(*this) += p; } mint operator-(const mint& p) const { return mint(*this) -= p; } mint operator*(const mint& p) const { return mint(*this) *= p; } mint operator/(const mint& p) const { return mint(*this) /= p; } bool operator==(const mint& p) const { return x == p.x; } bool operator!=(const mint& p) const { return x != p.x; } mint pow(int64_t n) const { mint res = 1, mul = x; while(n > 0){ if(n & 1) res *= mul; mul *= mul; n >>= 1; } return res; } mint inverse() const { return pow(mod - 2); } friend ostream& operator<<(ostream& os, const mint& p){ return os << p.x; } friend istream& operator>>(istream& is, mint& p){ int64_t val; is >> val; p = mint(val); return is; } }; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); ll N; cin >> N; mint ans = 0, sum = mint(N) * (N + 1) / 2; for(ll i = 1; i <= N; i *= 2){ mint a = mint(i * 2) * (i * 2 - 1) / 2 - mint(i) * (i - 1) / 2; mint d = mint(i * 2) * i; ll n = N / (i * 2); ll M = min(N, n * i * 2 + i - 1); ans += a * n + mint(n) * (n - 1) / 2 * d + (sum - mint(M) * (M + 1) / 2); } cout << ans << endl; return 0; }