In school we have learned how to do long addition with pen and paper in base 10. For example, to calculate 123+5678, we first write

   123
+ 5678
  ----

and then add the least-significant digits 3+8=11, which results in a carry:

    1
   123
+ 5678
  ----
     1

Next we will take the carry digit into account, and calculate 1+3+7=10, which again results in a carry, etc. Eventually we will have

   11
   123
+ 5678
  ----
  5801

Very good, 123+5678 = 5801. But what would happen if we did carryless addition in base 10? In essence, we will do long addition, but we will simply silently discard all carry digits, resulting in

   123
+ 5678
  ----
  5791

So the carryless sum in base 10 of 123 and 5678 is 5791. Your task is to write a program that calculates carryless sums in base 10.

Input

The first line contains a single number, a, and the second line contains a single number, b. The numbers consist of the digits 0–9. There are no leading zeroes (except if the number is "0").

Output

A single line containing c, which is the carryless sum of a and b in base 10. There are no leading zeroes (except if the number is "0").

Limits

The number of digits in a and b is between 1 and 10^6.

Example 1

Input:

12345
67890

Output:

79135

Example 2

Input:

123
5678

Output:

5791

Example 3

Input:

123
0

Output:

123

Example 4

Input:

1235
9860

Output:

95

Example 5

Input:

93
17

Output:

0

Example 6

Input:

98765434576890987657890876545776216383278432173246732846218462194210236487132054325019237621430
986127892147215879215198764567329173641207421467021765320342879234689325023476260293472175878238916437

Output:

986127880802649345005075311357195618317413704635453838566074615442041419233602647325426490887465537867