Divisible by 3 Again

Time Limit: 1 sec

**The Problem**

A number is divisible by 3 if the sum of all the digits is divisible by 3.

[If sum of all digit greater than 10 you can also sum the digits]

Example 1. 34164

Sum of all digits = 3 + 4 + 1 + 6 + 4 = 18

18 is divisible by 3, so 34164 is divisible by 3.

Sum of all digit = 3 + 4 + 1 + 6 + 4 = 18 => 1 + 8 = 9

9 is divisible by 3, so 34164 is divisible by 3.

Example 2. 16499205854376

Sum of all digits = 1 + 6 + 4 + 9 + 9 + 2 + 0 + 5 + 8 + 5 + 4 + 3 + 7 + 6 = 69

69 is divisible by 3, so 16499205854376 is divisible by 3.

Sum of all digits = 1 + 6 + 4 + 9 + 9 + 2 + 0 + 5 + 8 + 5 + 4 + 3 + 7 + 6 = 69 => 6 + 9 = 15 => 1 + 5 = 6

6 is divisible by 3, so 16499205854376 is divisible by 3.

Make a simple program that reads an integer number **N** and print "**YES**" if the number is divisible by 3 otherwise print "**NO**".

**The Input**

Input file contain a series of line, each line contain one integer number **N** (0<**N**<=**10**^{100}). Input is terminated by **EOF**.

**The Output**

For each line of input, print "**YES**" if the number is divisible by 3 otherwise print "**NO**" in a separate line.

**Sample Input**

16499205854376

10

123456789123456789123456789123456789

**Sample Output**

YES

NO

YES

Problem Setter: Shahin ShamS