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: Time Limit Exceeded)

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <numeric>
#include <random>
#include <vector>
#include <array>
#include <bitset>
#include <queue>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>

using namespace std;
using ll = long long;
using ull = unsigned long long;
constexpr ll TEN(int n) { return (n==0) ? 1 : 10*TEN(n-1); }
int bsr(int x) { return 31 - __builtin_clz(x); }

template<uint MD>
struct ModInt {
    uint v;
    ModInt() : v{0} {}
    ModInt(ll v) : v{normS(v%MD+MD)} {}
    static uint normS(const uint &x) {return (x<MD)?x:x-MD;};
    static ModInt make(const uint &x) {ModInt m; m.v = x; return m;}
    ModInt operator+(const ModInt &r) const {return make(normS(v+r.v));}
    ModInt operator-(const ModInt &r) const {return make(normS(v+MD-r.v));}
    ModInt operator*(const ModInt &r) const {return make((ull)v*r.v%MD);}
    ModInt& operator+=(const ModInt &r) {return *this=*this+r;}
    ModInt& operator-=(const ModInt &r) {return *this=*this-r;}
    ModInt& operator*=(const ModInt &r) {return *this=*this*r;}
    static ModInt inv(const ModInt &x) {
        return pow(ModInt(x), MD-2);
    }
};
using Mint = ModInt<TEN(9)+7>;

map<ll, Mint> dp;
Mint calc(ll N) {
    if (N == 0) return 1;
    if (dp.count(N)) return dp[N];
    Mint ans = 0;
    for (int i = 0; i <= 2; i++) {
        if (N < i) break;
        ans += calc((N-i)/2);
    }
    return ans;
}

int main() {
    ll n;
    cin >> n;
    cout << calc(n).v << endl;
    return 0;
}