Problem:
Score : 200 points

Problem Statement

An integer N is a multiple of 9 if and only if the sum of the digits in the decimal representation of N is a multiple of 9.

Determine whether N is a multiple of 9.

Constraints

0 \leq N < 10^{200000}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is a multiple of 9, print Yes; otherwise, print No.

Sample Input 1

123456789

Sample Output 1

Yes

The sum of these digits is 1+2+3+4+5+6+7+8+9=45, which is a multiple of 9, so 123456789 is a multiple of 9.

Sample Input 2

0

Sample Output 2

Yes

Sample Input 3

31415926535897932384626433832795028841971693993751058209749445923078164062862089986280

Sample Output 3

No

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>
#include <stdbool.h>

#define rep(i, n) for(int i=0; i<(n); ++i)
#define max  200001

//qsort(str, n, sizeof(int), up_comp);
int up_comp(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_comp(const void *a, const void *b){return *(int*)b - *(int*)a;}
int int_sort( const void * a , const void * b ) {return strcmp(( char * )a , ( char * )b );}
int count_digit(int a){int i=0;while(a!=0){a=a/10;i++;}return i;}
int euclid( int a, int b ){int temp;if(a<b){temp=a;a=b;b=temp;}if(b<1)return -1;if(a%b==0) return b;return euclid(b,a%b);}

int main(void){
    
    long long int n, sum=0;
    int i=10;
    
    scanf("%lld", &n);
    
    while (n>0) {
        sum+=n%i;
        n=n/10;
    }
    
    
    if (sum%9==0) {
        printf("YES");
    } else {
        printf("NO");
    }

    return 0;
}