diff --git a/RSA-encryption/Attack-LSBit-Oracle/README.md b/RSA-encryption/Attack-LSBit-Oracle/README.md
index 5fa3f85..0edda0b 100644
--- a/RSA-encryption/Attack-LSBit-Oracle/README.md
+++ b/RSA-encryption/Attack-LSBit-Oracle/README.md
@@ -91,8 +91,9 @@ Consider two messages 2\*m and 4\*m [\[1\]](https://crypto.stackexchange.com/que
 You must be wondering how condition 3 and 4 above hold true. You will understand better when you see the script implementing this:  
 ```python
 e = 65537
-upper_limit = N
+upper_limit = 1
 lower_limit = 0
+denominator = 1
 
 flag = ""
 i = 1
@@ -100,16 +101,21 @@ i = 1
 while i <= 1024:
     chosen_ct = long_to_bytes((bytes_to_long(flag_enc)*pow(2**i, e, N)) % N)
     output = _decrypt(chosen_ct)
+    delta = upper_limit - lower_limit
+    upper_limit *= 2
+    lower_limit *= 2
+    denominator *= 2
     if ord(output[-1]) == 0:
-        upper_limit = (upper_limit + lower_limit)/2
+        upper_limit -= delta
     elif ord(output[-1]) == 1:
-        lower_limit = (lower_limit + upper_limit)/2
+        lower_limit += delta
     else:
         throw Exception
     i += 1
 
 # Decrypted ciphertext
-print long_to_bytes(upper_limit)
+flag = upper_limit * N / denominator
+print long_to_bytes(flag)
 ```
 
 Now try understanding the conditions 3 and 4 using the above script:
@@ -119,7 +125,7 @@ Now try understanding the conditions 3 and 4 using the above script:
 
 It is somewhat similar to binary search algorithm, because our attack is doing the same (lowering the range of `m` until we get `m`), hence the complexity of this attack is **log<sub>2</sub>n**.
 
-Note that the above script messes up the last byte of the plaintext. I am trying to find out why, and will update the README as soon as I get it.
+Note that we keep track of the numerators and denominators separately, or else we end up with a messed up last byte from truncating along the way.
 
 I have written a script illustrating a server vulnerable to this attack, which you can find here: [encrypt.py](encrypt.py). You can also find the script I wrote to attack the above vulnerable service, which you can find here: [exploit.py](exploit.py)  
 
diff --git a/RSA-encryption/Attack-LSBit-Oracle/exploit.py b/RSA-encryption/Attack-LSBit-Oracle/exploit.py
index fe94d98..8dc3a3e 100644
--- a/RSA-encryption/Attack-LSBit-Oracle/exploit.py
+++ b/RSA-encryption/Attack-LSBit-Oracle/exploit.py
@@ -31,21 +31,25 @@ def _decrypt(ciphertext):
 print "\n\n"
 
 e = 65537
-upper_limit = N
+upper_limit = 1
 lower_limit = 0
+denominator = 1
 
 flag = ""
-i = 1
-while i <= 1034:
+for i in range(1, N.bit_length()+1):
     chosen_ct = long_to_bytes((bytes_to_long(flag_enc)*pow(2**i, e, N)) % N)
     output = _decrypt(chosen_ct)
+    delta = upper_limit - lower_limit
+    upper_limit *= 2
+    lower_limit *= 2
+    denominator *= 2
     if ord(output[-1]) == 0:
-        upper_limit = (upper_limit + lower_limit)/2
+        upper_limit -= delta
     elif ord(output[-1]) == 1:
-        lower_limit = (lower_limit + upper_limit)/2
+        lower_limit += delta
     else:
         break
         print "Unsuccessfull"
-    i += 1
 
-print "Flag : ", long_to_bytes(lower_limit)
+flag = N * lower_limit / denominator
+print "Flag : ", long_to_bytes(flag)
diff --git a/RSA-encryption/Attack-LSBit-Oracle/lsbitoracle.py b/RSA-encryption/Attack-LSBit-Oracle/lsbitoracle.py
index 5cf5fec..aa2c3f4 100644
--- a/RSA-encryption/Attack-LSBit-Oracle/lsbitoracle.py
+++ b/RSA-encryption/Attack-LSBit-Oracle/lsbitoracle.py
@@ -2,14 +2,12 @@
 from Crypto.PublicKey import RSA
 
 
-def lsbitoracle(flag_enc, _decrypt, e, N, upper_limit, lower_limit):
+def lsbitoracle(flag_enc, _decrypt, e, N):
     """
     Reference: https://crypto.stackexchange.com/questions/11053/rsa-least-significant-bit-oracle-attack
 
     Function implementing LSBit Oracle Attack
 
-    *Warning*: Function does not return the last byte of the final plaintext
-
     :parameters:
         flag_enc   : str
                     Ciphertext you want to decrypt
@@ -19,26 +17,27 @@ def lsbitoracle(flag_enc, _decrypt, e, N, upper_limit, lower_limit):
                     Public Key exponent
         N          : long
                     Public Key Modulus
-        upper_limit: long
-                    Maximum value of corresponding plaintext of flag_enc
-        lower_limit: long
-                    Minimum value of corresponding plaintext of flag_enc
 
-    Since the attack messes up with the last byte of the plaintext, lsbitoracle
-    function returns only flag[:-1]. It returns -1 in case of any Exception
+    Returns -1 in case of any Exception
     """
     flag = ""
     i = 1
-    while lower_limit < upper_limit:
+    lower_limit = 0
+    upper_limit = 1
+    denominator = 1
+    for i in range(1, N.bit_length()+1):
         chosen_ct = long_to_bytes((bytes_to_long(flag_enc)*pow(2**i, e, N)) % N)
         output = _decrypt(chosen_ct)
+        delta = upper_limit - lower_limit
+        upper_limit *= 2
+        lower_limit *= 2
+        denominator *= 2
         if ord(output[-1]) == 0:
-            upper_limit = (upper_limit + lower_limit)/2
+            upper_limit -= delta
         elif ord(output[-1]) == 1:
-            lower_limit = (lower_limit + upper_limit)/2
+            lower_limit += delta
         else:
             return -1
         i += 1
-    # clearing the last byte from the flag
-    flag = lower_limit & (~0xff)
+    flag = N * lower_limit / denominator
     return long_to_bytes(flag)