Trial Problem

Time limit1sMemory limit128 MB

Summary

Given n up to 2010, compute the final value of the triple loop that adds each floored s*k/i term modulo 2010.

Level: Medium7 of 10
Topics: Math, Number theory
Solved: No attempts yet

Problem

You are given an integer n. Run the following C code exactly as written. After all three nested loops finish, determine the value of the variable ret.

int ret = 0;
for(int s = 1; s<=n; s++) {
  for(int k = s; k<=n; k++) {
    for(int i = k; i<=n; i++) {
      ret = (ret+s*k/i)%2010;
    }
  }
}

Here s*k/i is C integer division (it discards the fractional part), and because ret is reduced modulo 2010 on every iteration, ret always stays between 0 and 2009 inclusive.

Input

The first and only line of input contains the integer n ( $1 \le n \le 2010$ ).

Output

Print the final value of the variable ret on a single line after the loop has fully executed.

Examples3

Example 1
Input
```
1
```
Expected output
```
1
```
Example 2
Input
```
2
```
Expected output
```
4
```
Example 3
Input
```
5
```
Expected output
```
48
```