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: Runtime Error)

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<stdbool.h>
#define e 1000000007

typedef  long long ll;

int int_cmp(const void *a, const void *b){ 
    const int *ia = (const int *)a;
    const int *ib = (const int *)b;
    return *ia  - *ib; 
} 


ll int_cmp1(const void *a, const void *b){ 
    const int *ia = (const int *)a;
    const int *ib = (const int *)b;
    return *ia  - *ib; 
} 


int max(int a,int b){
	if(a<b)
		a=b;
	return a;	
}

int min(int a,int b){
	if(a>b)
		a=b;
	return a;	
}

int absolu(int n){
	if(n<0)
		n*=-1;
	
	return n;	
}


void solve(){
   char s[10000000000000000000000000];
	int i;
	
	scanf("%s",s);
	ll ans=0,len=strlen(s);
	
	for(i=0 ; i<len ; i++){
		int val=s[i]-'0';
		ans+=val;
	}
	
	 if(ans%9==0)
		printf("YES\n");
	else
		printf("NO\n");		
}

 





int main(){
 /* int t;
	scanf("%d",&t);
	
	while(t--)*/
		solve();
	
	
	
	return 0;
}