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 <stdlib.h>
int main(void){
    // Your code here!
    char num[200000];
    int num_len;
    int num_add=0;
    int i;
    
    gets(num);
    // exception(0 is not calculated)
    if(num[0] == '0'){
        printf("No\n");
        return 0;
    }
    num_len = strlen(num);
    //printf("length:%d\n", num_len);
    if((num_len < 1) || (num_len > 100000)){
        printf("incorrect length\n");
        return -1;
    }
    
    for(i=0;i<num_len;i++){
        if((num[i] < '0') || (num[i] > '9')){
            printf("incorrect number:%d\n", i+1);
            return -1;
        }
        else{
            num_add += num[i]-0x30;
        }
    }
    
    if(num_add%9){
        printf("No\n");
    }
    else{
        printf("Yes\n");
    }
    
    return 0;
    
}