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: Runtime Error)

#include <bits/stdc++.h>

using namespace std;

using i64 = int64_t;

const i64 MOD = 1e9+7;

const i64 INF = 1e18+7;


signed main(){

    vector<int> aa;
    vector<int> bb(16003, -1);
    bb[0] = 0;
    bb[1] = 1;
    function<int(int)> f = [&](int x){
        if(bb[x] != -1)
            return bb[x];
        if(x & 1)
            return bb[x] = f(x / 2) + f(x / 2 + 1);
        else
            return bb[x] = f(x / 2);
    };

    for(int i = 0; i < 16003; ++i)
        f(i);
    vector<int> c(1, 0);
    for(auto& x : bb)
        c.emplace_back((c.back() + x) % MOD);

    i64 n;
    cin >> n;
    cout << c[n + 2] << endl;
}