SGU 105

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


找规律易知会出现011 011 ..... 011 的情况（0代表不可div,1代表可div) 推公式直接输出即可。

#include <iostream>
#include <cstdio>

using namespace std;

long long n;
int a[3] = {0,0,1};
int main()
{
    scanf("%I64d",&n);

    printf("%d
",n / 3 * 2 + a[n % 3] );
    return 0;
}