Problem:
Score : 600 points

Problem Statement

You are given a positive integer N.
Find the number of the pairs of integers u and v (0≦u,v≦N) such that there exist two non-negative integers a and b satisfying a xor b=u and a+b=v.
Here, xor denotes the bitwise exclusive OR.
Since it can be extremely large, compute the answer modulo 10^9+7.

Constraints

1≦N≦10^{18}

Input

The input is given from Standard Input in the following format:

N

Output

Print the number of the possible pairs of integers u and v, modulo 10^9+7.

Sample Input 1

3

Sample Output 1

5

The five possible pairs of u and v are:

u=0,v=0 (Let a=0,b=0, then 0 xor 0=0, 0+0=0.)

u=0,v=2 (Let a=1,b=1, then 1 xor 1=0, 1+1=2.）

u=1,v=1 (Let a=1,b=0, then 1 xor 0=1, 1+0=1.）

u=2,v=2 (Let a=2,b=0, then 2 xor 0=2, 2+0=2.）

u=3,v=3 (Let a=3,b=0, then 3 xor 0=3, 3+0=3.）

Sample Input 2

1422

Sample Output 2

52277

Sample Input 3

1000000000000000000

Sample Output 3

787014179

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using int64 = long long;

constexpr int DEBUG = 0;

constexpr int64 P = 1000000007;

struct FiniteField {
 private:
  int64 x;
  static inline int64 Normalize(int64 x) {
    if (0 <= x && x < P) return x; x %= P; if (x < 0) x += P; return x;
  }
 public:
  FiniteField(int64 x) : x(Normalize(x)) {}
  FiniteField() : x(0) {}
  int64 Value() const { return x; }
  inline FiniteField operator+(FiniteField o) const {
    FiniteField r(*this); r += o; return r;
  }
  inline FiniteField operator-(FiniteField o) const {
    FiniteField r(*this); r -= o; return r;
  }
  inline FiniteField operator* (FiniteField o) const {
    FiniteField r(*this); r *= o; return r;
  }
  inline FiniteField operator/ (FiniteField o) const {
    FiniteField r(*this); r /= o; return r;
  }
  inline void operator+= (FiniteField o) { x = (x + o.x) % P; }
  inline void operator-= (FiniteField o) { x = (x + P - o.x) % P; }
  inline void operator*= (FiniteField o) { x = (x * o.x) % P; }
  void operator/=(FiniteField o) {
    int64 p = P - 2; while (p) { if (p % 2) { *this *= o; } o *= o; p /= 2; }
  }
};

ostream& operator<<(ostream& s, const FiniteField& v) { s << v.Value(); return s; }

FiniteField Power(FiniteField x, int p) {
  if (p < 0) return FiniteField(1) / Power(x, -p);
  FiniteField value = 1;
  while (p) {
    if (p % 2) value *= x;
    x *= x;
    p /= 2;
  }
  return value;
}

template<typename T>
vector<vector<vector<T>>>
Make3DVector(int d1, int d2, int d3, T default_value) {
  return vector<vector<vector<T>>>(
      d1, vector<vector<T>>(d2, vector<T>(d3, default_value)));
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);

  int64 n;
  cin >> n;

  map<int64, FiniteField> dp;
  function<FiniteField(int64)> solve_dp_fn = [&](int64 x) -> FiniteField {
    if (dp.find(x) != dp.end()) {
      return dp[x];
    }
    if (x == 0) return 1;
    if (x == 1) return 2;
    auto v = solve_dp_fn((x - 2) / 2) + solve_dp_fn((x - 1) / 2) + solve_dp_fn(x / 2);
    dp[x] = v;
    return v;
  };
  cout << solve_dp_fn(n) << endl;
  // cout << ans << endl;
}