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)

#べき乗(MODあり)(繰り返し２乗法)
def pow(x, n, mod)
  ans = 1
  while n > 0
    ans = ans * x % mod if n.odd?
    x = x * x % mod
    n >>= 1
  end
  ans
end

L, R = gets.chomp.split(" ").map(&:to_i)
MOD = 10 ** 9 + 7

l_blen = L.bit_length
r_blen = R.bit_length
m = 0
((l_blen+1)...r_blen).each{|i|
  m += pow(3, i - 1, MOD)
}

def count_fraction(fraction, len, &b)
  wk = fraction
  ret, cnt = [0, 0]
  (1...len).each{|c|
    if b.call(wk) then
      ret = (2 * ret + pow(3, c - 1, MOD)) % MOD
      cnt += 1
    end
    wk >>= 1
  }
  (ret + pow(2, cnt, MOD)) % MOD
end

r = count_fraction(R, r_blen){|n| n.odd?}
if l_blen == r_blen then
  l = 0
else
  l = count_fraction(L, l_blen){|n| n.even?}
end

puts (m + l + r) % MOD