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 Ruby (2.3.3) (judge verdict: Wrong Answer)

L, R = gets.split.map(&:to_i)

# (L..R).each do |x|
#   (x..R).each do |y|
#     puts "#{"%#b" % y} % #{"%#b" % x} = #{y % x}, #{"%#b" % y} ^ #{"%#b" % x} = #{y ^ x}" if (y % x) == (y ^ x)
#   end
# end

# 二進でn桁：3**(n-1)個
sum = 0
(L.bit_length..R.bit_length).each do |n|
  sum += 3 ** (n-1)
end
(2**L.bit_length...L).each do |i|
  i.bit_length.times do |j|
    sum -= 2 if i[j] == 0
  end
end
(2**(R.bit_length+1)..R).each do |i|
  i.bit_length.times do |j|
    sum += 2 if i[j] == 0
  end
end
puts sum % (10**9+7)