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 Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

fn nth_bit(n: i64, bit: usize) -> i64 {
    (n >> bit) & 1
}

const MOD: i64 = 1000000000 + 7;

fn solve(n: i64) -> i64 {
    let digit = (n as f64).log2() as usize + 1;
    let mut dp = vec![vec![0; 3]; digit];
    dp[digit - 1][2] = 0;
    dp[digit - 1][1] = 1;
    dp[digit - 1][0] = 1;
    for i in (0..digit-1).rev() {
        let x = nth_bit(n, i);
            for j in 0..3 {
                for k in 0..3 {
                    let nj = j * 2 + x - k;
                    let i = i as usize;
                    let j = j as usize;
                    if nj >= 2 {
                        dp[i][2] = dp[i][2] % MOD + dp[i+1][j] % MOD;
                    } else if nj >= 0 {
                        dp[i][nj as usize] = dp[i][nj as usize] % MOD +  dp[i+1][j] % MOD;
                    }
                }
            }
    }

    (dp[0][0] % MOD + dp[0][1] % MOD + dp[0][2] % MOD) % MOD
}

fn main() {
    input! {
        n: i64,
    }
    println!("{}", solve(n));
}