#include <iostream>
using namespace std;
const int N = 1e3 + 7, mod = 1e9 + 7;
int f[N][N];
int main() {
int n;
cin >> n;
for (int i = 0; i <= n; i ++) {
f[i][0] = 1; // 容量为0时,前 i 个物品全不选也是一种方案
}
for (int i = 1; i <= n; i ++) {
for (int j = 0; j <= n; j ++) {
f[i][j] = f[i - 1][j] % mod; // 特殊 f[0][0] = 1
if (j >= i) f[i][j] = (f[i - 1][j] + f[i][j - i]) % mod;
}
}
cout << f[n][n] << endl;
}