#include #include #include #include using namespace std; using namespace __gnu_pbds; template using ordered_set = tree, rb_tree_tag, tree_order_statistics_node_update>; struct __INIT{__INIT(){cin.tie(0);ios::sync_with_stdio(false);cout<=0;i--) #define FOR(i,a,b) for(ll i=a;i<=ll(b);i++) #define FORD(i,a,b) for(ll i=a;i>=ll(b);i--) #define overload4(_1,_2,_3,_4,name,...) name #define overload3(_1,_2,_3,name,...) name #define rep1(n) for(ll i=0;i(a);) #define rrep4(i,a,b,c) for(ll i=(a)+((b)-(a)-1)/(c)*(c);i>=(a);i-=c) #define rrep(...) overload4(__VA_ARGS__,rrep4,rrep3,rrep2,rrep1)(__VA_ARGS__) #define ALL(v) v.begin(), v.end() #define endl "\n" #define fi first #define se second #define popcount(bit) __builtin_popcount(bit) #define popcountll(bit) __builtin_popcountll(bit) #define pb push_back #define eb emplace_back #define sz(x) ((ll)(x).size()) using namespace atcoder; using P = pair; using PL = pair; using Graph = vector>; typedef long long ll; template inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } inline void Yes(bool b = true) { cout << (b ? "Yes" : "No") << '\n'; } inline void YES(bool b = true) { cout << (b ? "YES" : "NO") << '\n'; } inline void OKNG(bool b = true) { cout << (b ? "OK" : "NG") << '\n'; } const int dx[4] = {1, 0, -1, 0}; const int dy[4] = {0, 1, 0, -1}; const int fx[8] = {0, 1, 1, 1, 0, -1, -1, -1}; const int fy[8] = {1, 1, 0, -1, -1, -1, 0, 1}; template const auto INF = numeric_limits::max()/2; using mint = modint1000000007; string to_binary(ll N, int digit = 0) { string res = ""; while (N) { if (N & 1) res = "1" + res; else res = "0" + res; N >>= 1; } if (0 < digit) { int n = digit - res.length(); rep(i, 0, n) res = "0" + res; } return res; } int main() { ll n; cin >> n; vector s; while(n > 0){ s.push_back(n % 2); n /= 2; } reverse(ALL(s)); n = s.size(); mint com[n + 1][n + 2][2] = {0}; mint sum[n + 1][n + 2][2] = {0}; com[0][0][0] = 1; rep(i, n)rep(j, n+1)rep(k, 2){ ll nd = s[i]; rep(d, 2){ ll nk = k; ll nj = j; if(k == 0 && d > nd)continue; if(d < nd)nk = 1; if(d == 1)nj++; com[i+1][nj][nk] += com[i][j][k]; sum[i+1][nj][nk] += sum[i][j][k] * 2 + com[i][j][k] * d; } } mint ans = 0; rep(i, n+2)rep(j, 2)ans += sum[n][i][j] * i; cout << ans.val() << endl; }