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 <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <utility>
#include <queue>
#include <set>
#include <map>

using namespace std;
typedef  long long ll;
typedef pair<int,int> PII;
typedef vector<int> VI;
typedef vector<VI> VVI;
#define  MP make_pair
#define  PB push_back
#define inf  1000000007
#define rep(i,n) for(int i=0;i<(int)(n);++i)

template<typename A, size_t N, typename T>
void Fill(A (&array)[N], const T &val){
    std::fill( (T*)array, (T*)(array+N), val );
}

ll mod = inf;
int x[60];
ll dp[60],dp2[60],dp3[60];

int main(){
    ll n,m;
    cin >> n;
    m = n;
    int c = 0;
    for(int i=0;i<60;i++){
        if(m%2!=0){
            x[i] = 1;
            c = i;
        }else{
            x[i] = 0;
        }
        m/=2;
    }
    if(x[0]==0){
        dp[0] = 1;
        dp2[0] = 3;
        dp3[0] = 3;
    }else{
        dp[0] = 2;
        dp2[0] = 3;
        dp3[0] = 3;
    }
    for(int i=1;i<=c;i++){
        if(x[i]==0){
            dp[i] = dp[i-1];
            dp2[i] = (dp3[i-1]+dp2[i-1]+dp[i-1])%mod;
        }else{
            dp[i] = (dp[i-1]+dp2[i-1])%mod;
            dp2[i] = (2*dp3[i-1]+dp2[i-1])%mod;
        }
        dp3[i] = (3*dp3[i-1])%mod;
    }
    cout << dp[c] << endl;

    return 0;
}