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;
long long mod=1e9+7;
long long no[180],ans[180];
map<long long,long long> mp;
long long f(long long k)
{
    if (mp[k]) return mp[k];
    for(int i=1;i<=177;i++)
        if (k==no[i]) return mp[k]=ans[i];
    if (k%2==0) return mp[k]=(2*f(k/2-1)+f(k/2))%mod;
    return mp[k]=(f(k/2-1)+2*f(k/2))%mod;
}
int main()
{
     long long n;
     cin>>n;
     no[1]=0;no[2]=1;no[3]=2;
     ans[1]=1;ans[2]=2;ans[3]=4;
     for(int i=2;i<=59;i++)
     {
          no[3*i-2]=(1ll<<i)-2;
          no[3*i-1]=no[3*i-2]+1;
          no[3*i]=no[3*i-1]+1;
          ans[3*i-2]=(2*ans[3*i-5]+ans[3*i-4])%mod;
          ans[3*i-1]=(ans[3*i-5]+2*ans[3*i-4])%mod;
          ans[3*i]=(ans[3*i-1]+i+1)%mod;
     }
     cout<<f(n)<<endl;
     return 0;
}