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 (1.15.1) (judge verdict: Accepted)

fn run() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let mut it = s.trim().split_whitespace();
    let l: usize = it.next().unwrap().parse().unwrap();
    let r: usize = it.next().unwrap().parse().unwrap();
    const MOD: u64 = 1_000_000_007;
    let mut dp = [[0u64; 2]; 2];
    for shift in (0..60).rev() {
        let (x, y) = ((l >> shift) & 1, (r >> shift) & 1);
        let mut next = [[0; 2]; 2];
        for i in 0..2 {
            for j in 0..2 {
                let v = dp[i][j] % MOD;
                if x == 0 || i == 1 {
                    next[i][j | y] += v;
                }
                if (x == 0 || i == 1) && (y == 1 || j == 1) {
                    next[i][j] += v;
                }
                if y == 1 || j == 1 {
                    next[i | if x == 0 {1} else {0}][j] += v;
                }
            }
        }
        if 1 >= (l >> shift) && 1 <= (r >> shift) {
            next[if x == 0 {1} else {0}][if 1 < (r >> shift) {1} else {0}] += 1;
        }
        dp = next;
    }
    let mut ans = 0;
    for i in 0..2 {
        for j in 0..2 {
            ans += dp[i][j];
        }
    }
    ans %= MOD;
    println!("{}", ans);
}

fn main() {
    run();
}