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 <iostream>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <queue>
#include <stack>
#include <list>
#include <utility>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
#include <climits>
#include <map>
#include <unordered_map>
#include <string.h>
#include<iomanip>
#include <iterator>

#define all(a) a.begin(),a.end()
#define VV(T) vector<vector< T > >
#define rep(i,N) for(int i = 0; i <N; i++)
#define fill(x,y) memset(x,y,sizeof(x))
#define name(x) #x
using namespace std;


const int INF = INT_MAX;
const int MOD = pow(10,9)+7;
int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

typedef long long ll,LL;
typedef vector<int> vint;
typedef vector<vector<int> > vvint;
typedef vector<long long> vll, vLL;
typedef vector<vector<long long> > vvll, vvLL;
typedef pair<int,int> pii;
typedef pair<LL, LL> pLL;


// vector入力
template<typename T>
istream& operator >> (istream& is, vector<T>& vec){
    for(T& x: vec) is >> x;
    return is;
}
// vector出力
template<typename T>
ostream& operator << (ostream& os, vector<T>& vec){
    for(int i=0; i<vec.size(); i++){
        os << vec[i] << ( i+1 == vec.size() ? "" : " " );
    }
    return os;
}

map<ll,ll> dp;

ll filldp(ll n){
    
    if (dp.count(n)) return dp[n];

    ll res = (filldp(n/2) + filldp((n-1)/2)) %MOD;
    res = (res + filldp((n-2)/2)) %MOD;
    
    dp[n] = res;
    return res;
    
}


int main() {
    ll N;
    cin >> N;
    
    dp[0] = 1;
    dp[1] = 2;
    
    ll res = filldp(N);
    
    cout << res << endl;
    
    return 0;
}