Problem:
Score : 600 points

Problem Statement

Given are integers L and R. Find the number, modulo 10^9 + 7, of pairs of integers (x, y) (L \leq x \leq y \leq R) such that the remainder when y is divided by x is equal to y \mbox{ XOR } x.

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

1 \leq L \leq R \leq 10^{18}

Input

Input is given from Standard Input in the following format:

L R

Output

Print the number of pairs of integers (x, y) (L \leq x \leq y \leq R) satisfying the condition, modulo 10^9 + 7.

Sample Input 1

2 3

Sample Output 1

3

Three pairs satisfy the condition: (2, 2), (2, 3), and (3, 3).

Sample Input 2

10 100

Sample Output 2

604

Sample Input 3

1 1000000000000000000

Sample Output 3

68038601

Be sure to compute the number modulo 10^9 + 7.

Solution in Rust (1.42.0) (judge verdict: Accepted)

fn run() {
    proconio::input! {
        l: usize,
        r: usize,
    }
    const MOD: u64 = 1_000_000_007;
    let mut dp = [[0u64; 2]; 2];
    dp[0][0] = 0;
    for i in (0..60).rev() {
        let mut next = [[0; 2]; 2];
        if l >> i <= 1 && r >> i >= 1 {
            let a = 1 ^ (l >> i & 1);
            let b = (r >> i > 1) as usize;
            next[a][b] += 1;
        }
        for (a, dp) in dp.iter().enumerate() {
            for (b, &v) in dp.iter().enumerate() {
                for &(x, y) in [(0, 0), (0, 1), (1, 1)].iter() {
                    if a == 0 && l >> i & 1 > x {
                        continue;
                    }
                    if b == 0 && r >> i & 1 < y {
                        continue;
                    }
                    let a = (a == 1 || x > (l >> i) & 1) as usize;
                    let b = (b == 1 || y < (r >> i) & 1) as usize;
                    next[a][b] += v;
                }
            }
        }
        next.iter_mut().flatten().for_each(|p| *p %= MOD);
        dp = next;
    }
    let ans = dp.iter().flatten().sum::<u64>() % MOD;
    println!("{}", ans);
}

fn main() {
    run();
}