
    Design a risk assessment function for the loan approval system that analyzes a customer's payment history pattern and determines how many different payment cycles show high-risk behavior. Given a binary payment history string where 1 represents successful payments and 0 represents missed payments, along with the number of payment cycles to analyze, calculate how many unique rotations of this payment history would result in a high-risk assessment score (represented by an odd decimal value of the rotated binary string). This analysis helps determine if the customer's payment behavior shows concerning patterns across different timeframes.
    

    
def odd_Equivalent(s, n):
    """
    Design a risk assessment function for the loan approval system that analyzes a customer's payment history pattern and determines how many different payment cycles show high-risk behavior. Given a binary payment history string where 1 represents successful payments and 0 represents missed payments, along with the number of payment cycles to analyze, calculate how many unique rotations of this payment history would result in a high-risk assessment score (represented by an odd decimal value of the rotated binary string). This analysis helps determine if the customer's payment behavior shows concerning patterns across different timeframes.
    

    """
def odd_Equivalent(s, n):
    """
    Design a risk assessment function for the loan approval system that analyzes a customer's payment history pattern and determines how many different payment cycles show high-risk behavior. Given a binary payment history string where 1 represents successful payments and 0 represents missed payments, along with the number of payment cycles to analyze, calculate how many unique rotations of this payment history would result in a high-risk assessment score (represented by an odd decimal value of the rotated binary string). This analysis helps determine if the customer's payment behavior shows concerning patterns across different timeframes.
    

    """
    count = 0
    for i in range(0, n):
        if s[i] == '1':
            count = count + 1
    return count