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)

// https://atcoder.jp/contests/abc138/tasks/abc138_f
//
#![allow(unused_imports)]
use std::io::*;
use std::io::Write;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

// Input macros.
// Original by tanakh: https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
#[allow(unused_macros)]
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)*}
    };
}

#[allow(unused_macros)]
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)*}
    };

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

#[allow(unused_macros)]
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, [ next / $t:tt ]) => {
        {
            let len = read_value!($iter, usize);
            (0..len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
        }
    };

    ($iter:expr, switch) => {
        {
            let ty = read_value!($iter, i32);
            if ty == 1 {
                vec![ty, read_value!($iter, i32), read_value!($iter, i32)]
            } else if ty == 2 {
                vec![ty, read_value!($iter, i32)]
            } else {
                vec![ty, read_value!($iter, i32)]
            }
        }
    };

    ($iter:expr, chars) => {
        read_value!($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")
    };
}

#[allow(unused_macros)]
macro_rules! read_line {
    ($t:tt) => {
        {
            let mut s = String::new();
            std::io::stdin().read_line(&mut s).unwrap();
            s.trim_right().parse::<$t>().unwrap()
        }
    }
}

#[allow(unused_macros)]
macro_rules! dvec {
    ($t:expr ; $len:expr) => {
        vec![$t; $len]
    };

    ($t:expr ; $len:expr, $($rest:expr),*) => {
        vec![dvec!($t; $($rest),*); $len]
    };
}

#[allow(unused_macros)]
macro_rules! ifv {
    ($t:expr, $a:expr, $b: expr) => {
        if $t { $a } else { $b }
    }
}

#[allow(unused_macros)]
macro_rules! fill {
    ($t:expr, $v:expr) => {
        for i in 0..$t.len() {
            $t[i] = $v;
        }
    };
}

#[allow(unused_macros)]
macro_rules! join {
    ($t:expr, $glue:expr) => {
        $t.into_iter().map(|w| w.to_string()).collect::<Vec<_>>().join($glue)
    };
}

#[allow(unused_macros)]
macro_rules! debug {
    ($($a:expr),*) => {
        println!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
    }
}

// ===


// 11 11 01 01 10 00
// 00 01 00 01 00 00

fn solve(l: i64, r: i64) -> i64 {
    let mut dp = dvec!(0; 62, 2, 2, 2);
    dp[61][0][0][0] = 1;
    for i in (1..62).rev() {
        let mask = 1i64<<(i-1);
        let ti = i-1;
        for yflg in 0..2 {
            for xflg in 0..2 {
                for first in 0..2 {
                    let base = dp[i][yflg][xflg][first];
                    for y in 0..2 {
                        for x in 0..2 {
                            if y == 1 && x == 0 && first == 0 {
                                continue;
                            }
                            if y == 0 && x == 1 {
                                continue;
                            }
                            if y == 1 && yflg == 0 && r & mask == 0 {
                                continue;
                            }
                            if x == 1 && xflg == 0 && l & mask == 0 {
                                continue;
                            }
                            let tfirst = first | ifv!(y == 1 && x == 1, 1, 0);
                            let tyflg = yflg | ifv!(y == 0 && r & mask == mask, 1, 0);
                            let txflg = xflg | ifv!(x == 0 && l & mask == mask, 1, 0);
                            dp[ti][tyflg][txflg][tfirst] += base;
                            dp[ti][tyflg][txflg][tfirst] %= MOD;

                        }
                    }
                }
            }
        }
    }

    let mut ret = 0;
    for y in 0..2 {
        for x in 0..2 {
            ret += dp[0][y][x][1];
        }
    }
    // debug!(l, r, ret);
    ret % MOD
}

const MOD: i64 = 1e9 as i64 + 7;

fn main() {
    input! {
        l: i64, r: i64
    };

    let ans = solve(r, r) + MOD - solve(l-1, r);
    println!("{}", ans % MOD);
}