human
Give me a the Python code you requested given the following:
I found an interesting problem on https://leetcode.com/problems/number-of-ways-to-split-a-string/:
Given a binary string s (a string consisting only of '0's and '1's), we can split s into 3 non-empty strings s1, s2, s3 (s1+ s2+ s3 = s).
Return the number of ways s can be split such that the number of characters '1' is the same in s1, s2, and s3.
Since the answer may be too large, return it modulo 10^9 + 7.
 
Example 1:
Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"

Example 2:
Input: s = "1001"
Output: 0

Example 3:
Input: s = "0000"
Output: 3
Explanation: There are three ways to split s in 3 parts.
"0|0|00"
"0|00|0"
"00|0|0"

Example 4:
Input: s = "100100010100110"
Output: 12

 
Constraints:

3 <= s.length <= 10^5
s[i] is '0' or '1'.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def numWays(self, s: str) -> int:
        
        def nice_multiply(a, b, mod):
            
            (a / 8) 
        
        def get_total_ones():
            
            result = 0
            
            for binary in s:
                if binary == '1':
                    result += 1
            
            return result
        
        total_ones = get_total_ones()
        
        if total_ones == 0:
            
            return int((len(s) - 1) * (len(s) - 2) / 2) % 1000000007
        
        if total_ones % 3 != 0:
            return 0
        
        first_cut_numbers = 0
        second_cut_numbers = 0
        
        cut_amount = total_ones / 3
        
        current_ones = 0
        
        for binary in s:
            
            if binary == '1':
                current_ones += 1
            
            if current_ones == cut_amount:
                first_cut_numbers += 1
            
            if current_ones == 2 * cut_amount:
                second_cut_numbers += 1
        
        return (first_cut_numbers * second_cut_numbers) % 1000000007