#include <iostream>
#include <cmath>
using namespace std;

typedef long long ll;

int main()
{
	ll N;
	cin >> N;

	ll a = N / 2 % 1000007;
	ll b = (int)ceil((double)N / 2) % 1000007;

	cout << ((a + 1) * b + a) % 1000007 << endl;

	return 0;
}