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 <stdlib.h>
#include <string.h>

#define rep(i, n) for (int i = 0; i < n; i = i + 1)
#define array(N, t) (t*)calloc(N, sizeof(t))
#define ll long long
#define INF ((1LL << 62) - (1LL << 31))
#define max(p, q)((p) > (q) ? (p) : (q))
#define min(p, q)((p) < (q) ? (p) : (q))

int main() {
  char N[20010];
  scanf("%s", N);

  int len = strlen(N);

  int sum = 0;
  rep(i, len) {
    sum += (N[i] - '0');
  }

  if (sum % 9 == 0) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }

  return 0;
}