#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int n;
int f[N][N];
int main() {
    cin >> n;
    f[0][0]=1; 
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= i; j ++ )
            f[i][j] = (f[i - 1][j - 1] + f[i - j][j]) % mod;
//以最小值是否为1来分类，如果是1，那么数量和总和都减去，如果不是，每个数字都减1
    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = (res + f[n][i]) % mod;
    cout << res << endl;
    return 0;
}
//f[i][j]表示总和为i，总个数为j的方案数