XOR solution explanation needed.

[omm]

Question:
For a given list of numbers, only one digit is repeated odd number of times. Pass the list through function and return the number which is repeated odd number of times.

I have a solution for it. But, I found another solution by other user which I did not understand. Here is the code

def find_it(seq):
    result = 0
    for x in seq:
        result ^= x
    return result

find_it([1,2,3,1,2,3,1])

XOR seems to be implemented but, how is the only odd number being detected.

More explanation:
I have tried printing the binary values of result and x before and after statement result ^= x to understand what is happening but, could not.
Any other resources that you can point me to get better understanding?

 For value 1 , Before ^x Operation. bin(result) =  0b0 , bin(x) =  0b1
 For value 1 , After ^x Operation. bin(result) =  0b1 , bin(x) =  0b1
Result =  1 

 For value 2 , Before ^x Operation. bin(result) =  0b1 , bin(x) =  0b10
 For value 2 , After ^x Operation. bin(result) =  0b11 , bin(x) =  0b10
Result =  3 

 For value 3 , Before ^x Operation. bin(result) =  0b11 , bin(x) =  0b11
 For value 3 , After ^x Operation. bin(result) =  0b0 , bin(x) =  0b11
Result =  0 

 For value 1 , Before ^x Operation. bin(result) =  0b0 , bin(x) =  0b1
 For value 1 , After ^x Operation. bin(result) =  0b1 , bin(x) =  0b1
Result =  1 

 For value 2 , Before ^x Operation. bin(result) =  0b1 , bin(x) =  0b10
 For value 2 , After ^x Operation. bin(result) =  0b11 , bin(x) =  0b10
Result =  3 

 For value 3 , Before ^x Operation. bin(result) =  0b11 , bin(x) =  0b11
 For value 3 , After ^x Operation. bin(result) =  0b0 , bin(x) =  0b11
Result =  0 

 For value 1 , Before ^x Operation. bin(result) =  0b0 , bin(x) =  0b1
 For value 1 , After ^x Operation. bin(result) =  0b1 , bin(x) =  0b1
Result =  1 

 For value 4 , Before ^x Operation. bin(result) =  0b1 , bin(x) =  0b100
 For value 4 , After ^x Operation. bin(result) =  0b101 , bin(x) =  0b100
Result =  5 

 For value 4 , Before ^x Operation. bin(result) =  0b101 , bin(x) =  0b100
 For value 4 , After ^x Operation. bin(result) =  0b1 , bin(x) =  0b100
Result =  1 

1

Process finished with exit code 0

Thanks,
Om