SGU 105 div.3 找规律

There is sequence 1, 12, 123, 1234, ..., 12345678910, ... . Given first N elements of that sequence. You must determine amount of numbers in it that are divisible by 3.

Input

Input contains N (1<=N<=2³¹ - 1).

Output

Write answer to the output.

Sample Input

4

Sample Output

2

题目大意：第个序列为1234...n，求前n项有几个数能被3整除题解：有规律

#include <stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    printf(*+(n%==?:));
    ;
}