#include using namespace std; //* ATCODER #include using namespace atcoder; typedef modint998244353 mint; //*/ /* BOOST MULTIPRECISION #include using namespace boost::multiprecision; //*/ typedef long long ll; #define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++) #define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--) template bool chmin(T &a, const T &b) { if (a <= b) return false; a = b; return true; } template bool chmax(T &a, const T &b) { if (a >= b) return false; a = b; return true; } template T max(vector &a){ assert(!a.empty()); T ret = a[0]; for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]); return ret; } template T min(vector &a){ assert(!a.empty()); T ret = a[0]; for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]); return ret; } template T sum(vector &a){ T ret = 0; for (int i=0; i<(int)a.size(); i++) ret += a[i]; return ret; } //defmodfact const int COMinitMAX = 998244; mint fact[COMinitMAX+1], factinv[COMinitMAX+1]; void modfact(){ fact[0] = 1; for (int i=1; i<=COMinitMAX; i++){ fact[i] = fact[i-1] * i; } factinv[COMinitMAX] = fact[COMinitMAX].inv(); for (int i=COMinitMAX-1; i>=0; i--){ factinv[i] = factinv[i+1] * (i+1); } } mint cmb(int a, int b){ if (a, pair> mp; // kosu, ans pair solve(ll m, int k){ if (m < 0) return pair(0, 0); chmin(m, (1LL << k+1)); if (mp.find(pair(m, k)) != mp.end()){ return mp[pair(m, k)]; } //cout << m << ' ' << k << endl; if (k == 0){ if (m == 0){ return pair(0, 0); }else if(m == 1){ return pair(2, 2); }else if(m >= 2){ return pair(4, 4 + inv2); } }else{ pair ret1 = solve(m, k-1); pair ret2 = solve(m - (1LL << k), k-1); pair ret =\ pair(ret1.first * 2 + ret2.first * 2,\ ret1.second * 2\ + ret2.second * inv2.pow(1LL << k)\ + ret2.second * mint(2).pow(1LL << k)); mp[pair(m, k)] = ret; return ret; } } int main(){ ll k = 0; ll m; cin >> m; while((1LL << k) < m){ k++; } if (k == 0){ cout << 1 << '\n'; return 0; } mint teisu1 = mint(2).pow((1LL << k) - 2) * inv2.pow((1LL << k-1)); mint ans = mint(2).pow(1LL << k-1) - 1; //cout << ans.val() << '\n'; ans += mint(2).pow((1LL << k-1) - (m - (1LL << k-1))) * (mint(2).pow(m - (1LL << k-1)) - 1); //cout << ans.val() << '\n'; //cout << solve(m - (1LL << k-1), k-1).second.val() << '\n'; ans += solve(m - (1LL << k-1), k-1).second * teisu1; cout << ans.val() << '\n'; }