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/***************************************************
* Problem Name : AE00 - Rectangles.cpp
* Problem Link : https://www.spoj.com/problems/AE00/en/
* OJ : Spoj
* Verdict : AC
* Date : 2019-02-08
* Problem Type : adHoc
* Author Name : Saikat Sharma
* University : CSE, MBSTU
***************************************************/
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<climits>
#include<cstring>
#include<string>
#include<sstream>
#include<cmath>
#include<vector>
#include<queue>
#include<cstdlib>
#include<deque>
#include<stack>
#include<map>
#include<set>
#define __FastIO ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#define __FileRead freopen ("/home/saikat/Desktop/logN/input.txt", "r", stdin);
#define __FileWrite freopen ("/home/saikat/Desktop/logN/output.txt", "w", stdout);
#define SET(a,v) memset(a,v,sizeof(a))
#define pii pair<int,int>
#define pll pair <ll, ll>
#define debug cout <<"#########\n";
#define nl cout << "\n";
#define sp cout << " ";
#define sl(n) scanf("%lld", &n)
#define sf(n) scanf("%lf", &n)
#define si(n) scanf("%d", &n)
#define ss(n) scanf("%s", n)
#define pf(n) scanf("%d", n)
#define pfl(n) scanf("%lld", n)
#define all(v) v.begin(), v.end()
#define Pow2(x) ((x)*(x))
#define Mod(x, m) ((((x) % (m)) + (m)) % (m))
#define Max3(a, b, c) max(a, max(b, c))
#define Min3(a, b, c) min(a, min(b, c))
#define pb push_back
#define mk make_pair
#define MAX 100005
#define INF 1000000000
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <typename T>
std::string NumberToString ( T Number ) {
std::ostringstream ss;
ss << Number;
return ss.str();
}
ll lcm (ll a, ll b) {
return a * b / __gcd (a, b);
}
/************************************ Code Start Here ******************************************************/
int check (ll n) {
ll sq = sqrt (n);
if (sq * sq == n) return 1;
else return 0;
}
int main () {
ll N;
while (scanf ("%lld", &N) == 1) {
ll sq = sqrt (N), ans = 0;
for (int i = 1; i <= sq; i++) {
int cnt = 0;
for (int j = i; j <= N; j++) {
if (i * j <= N) cnt++;
}
ans += cnt;
}
printf ("%lld\n", ans);
}
return 0;
}
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