Bit erase

You are given a bit string with n characters. In each step, you may erase two adjacent bits that are different. How many ways there are to erase all bits?

For example, when the bit string is 101100, there are 5 possible ways:

• \underline{10}1100 \rightarrow 1\underline{10}0 \rightarrow \underline{10} \rightarrow (empty)
• 1\underline{01}100 \rightarrow 1\underline{10}0 \rightarrow \underline{10} \rightarrow (empty)
• 101\underline{10}0 \rightarrow \underline{10}10 \rightarrow \underline{10} \rightarrow (empty)
• 101\underline{10}0 \rightarrow 1\underline{01}0 \rightarrow \underline{10} \rightarrow (empty)
• 101\underline{10}0 \rightarrow 10\underline{10} \rightarrow \underline{10} \rightarrow (empty)

You may assume that 1 \le n \le 30. The algorithm should be efficient in all these cases.

In a file biterase.py, implement a function count that returns the number of ways to erase all bits.

def count(s):
    # TODO

if __name__ == "__main__":
    print(count("1001")) # 2
    print(count("1100")) # 1
    print(count("101100")) # 5
    print(count("11001")) # 0
    print(count("01110100100110")) # 6027