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#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
const int MOD = 1e9 + 7;
int n;
int f[N][N];
int main() {
cin >> n;
for (int i = 1; i <= n; i++) f[i][0] = 1;
for (int i = 1; i <= n; i++) // 枚举每个物品，物品的序号与物品的体积是相等的，都是i
for (int j = 1; j <= n; j++) { // 枚举背包容量j
if (j >= i) // ① 背包容量j>=当前体积i,可以选择当前数字
f[i][j] = (f[i][j - i] + f[i - 1][j]) % MOD;
else
f[i][j] = f[i - 1][j] % MOD; // ② 放弃当前数字
}
cout << f[n][n] << endl;
return 0;
