#include <atcoder/all>
#include <bits/stdc++.h>
#define rep(i, a, b) for (ll i = (ll)(a); i < (ll)(b); i++)
using namespace atcoder;
using namespace std;

typedef long long ll;

template <class T, class S = T> struct AllPairOperationSum {
    // https://shakayami.hatenablog.com/entry/2021/01/01/044946
    // Σ op(a[i], a[j]) (i < j)
    ll n;
    vector<T> a;
    AllPairOperationSum() {}
    AllPairOperationSum(vector<T> a) : a(a), n(a.size()) {}
    S add() {
        // A + B
        // O(N)
        S ret = 0;
        rep(i, 0, n) ret += a[i];
        ret *= (n - 1);
        return ret;
    }
    S sub() {
        // A - B
        // O(N)
        S ret = 0;
        rep(i, 0, n) ret += (S)(n - 1 - i * 2) * a[i];
        return ret;
    }
    S mul() {
        // A * B
        // O(N)
        S ret = 0;
        vector<S> s(n);
        s[0] = a[0];
        rep(i, 1, n) {
            s[i] = s[i - 1] + a[i];
            ret += s[i - 1] * a[i];
        }
        return ret;
    }
    S div() {
        // A / B (有理数mod)
        // O(N log maxA)
        S ret = 0;
        vector<S> invs(n);
        reverse(a.begin(), a.end());
        invs[0] = (S)1 / a[0];
        rep(i, 1, n) invs[i] = invs[i - 1] + (S)1 / a[i];
        rep(i, 1, n) ret += invs[i - 1] * a[i];
        reverse(a.begin(), a.end());
        return ret;
    }
    T floor() {
        // floor(A / B)
        // O(N √maxA log maxA)
        T maxa = *max_element(a.begin(), a.end());
        T sqrtmaxa = sqrt(maxa);
        vector<T> v(sqrtmaxa, 0);
        fenwick_tree<T> ft(maxa + 1);
        T ret = 0;
        rep(i, 0, n) {
            if (a[i] < sqrtmaxa) {
                ret += v[a[i]];
            } else {
                for (T k = 0; k < maxa + 1; k += a[i]) {
                    ret += (k / a[i]) * ft.sum(k, std::min(k + a[i], maxa + 1));
                }
            }
            ft.add(a[i], 1);
            for (T j = 1; j < sqrtmaxa; j++) {
                v[j] += a[i] / j;
            }
        }
        return ret;
    }
    T mod() {
        // A % B
        // O(N √maxA log maxA)
        T maxa = *max_element(a.begin(), a.end());
        T sqrtmaxa = sqrt(maxa);
        vector<T> v(sqrtmaxa, 0);
        fenwick_tree<T> ft(maxa + 1);
        T ret = 0;
        rep(i, 0, n) {
            if (a[i] < sqrtmaxa) {
                ret += v[a[i]] * a[i];
            } else {
                for (T k = 0; k < maxa + 1; k += a[i]) {
                    ret += (k / a[i]) *
                           ft.sum(k, std::min(k + a[i], maxa + 1)) * a[i];
                }
            }
            ft.add(a[i], 1);
            for (T j = 1; j < sqrtmaxa; j++) {
                v[j] += a[i] / j;
            }
        }
        ret *= -1;
        rep(i, 0, n) ret += (n - 1 - i) * a[i];
        return ret;
    }
    S error_sq() {
        // (A - B)^2
        // O(N)
        S ret = 0;
        rep(i, 0, n) ret += (S)a[i] * a[i];
        ret *= (n - 1);
        ret += mul() * (-2);
        return ret;
    }
    S max() {
        // max(A, B)
        // O(N)
        S ret = 0;
        vector<T> b = a;
        sort(b.begin(), b.end());
        rep(i, 0, n) ret += (S)i * b[i];
        return ret;
    }
    S min() {
        // min(A, B)
        // O(N)
        S ret = 0;
        vector<T> b = a;
        sort(b.rbegin(), b.rend());
        rep(i, 0, n) ret += (S)i * b[i];
        return ret;
    }
    S error() {
        // |A - B|
        // O(N)
        S ret = 0;
        vector<T> b = a;
        sort(b.rbegin(), b.rend());
        rep(i, 0, n) ret += (S)(n - 1 - i * 2) * b[i];
        return ret;
    }
    S xor_() {
        // A xor B
        // O(N log maxA)
        T mx = *max_element(a.begin(), a.end());
        S ret = 0;
        int base = 0;
        while (mx >> base) {
            vector<ll> cnt(2);
            rep(i, 0, n) cnt[(a[i] >> base) & 1]++;
            ret += (S)(1LL << base) * (cnt[0] * cnt[1]);
            base++;
        }
        return ret;
    }
    S and_() {
        // A & B
        // O(N log maxA)
        T mx = *max_element(a.begin(), a.end());
        S ret = 0;
        int base = 0;
        while (mx >> base) {
            vector<ll> cnt(2);
            rep(i, 0, n) cnt[(a[i] >> base) & 1]++;
            ret += (S)(1LL << base) * (cnt[1] * (cnt[1] - 1) / 2);
            base++;
        }
        return ret;
    }
    S or_() {
        // A | B
        // O(N log maxA)
        T mx = *max_element(a.begin(), a.end());
        S ret = 0;
        int base = 0;
        while (mx >> base) {
            vector<ll> cnt(2);
            rep(i, 0, n) cnt[(a[i] >> base) & 1]++;
            ret += (S)(1LL << base) *
                   (n * (n - 1) / 2 - cnt[0] * (cnt[0] - 1) / 2);
            base++;
        }
        return ret;
    }
    S lcm() {
        // lcm(A, B)
        // O(maxA log maxA)
        T m = *max_element(a.begin(), a.end());
        vector<S> w(m + 1);
        rep(i, 1, m + 1) {
            w[i] += (S)1 / i;
            int di = i + i;
            while (di <= m) {
                w[di] -= w[i];
                di += i;
            }
        }
        vector<ll> b(m + 1);
        rep(i, 0, n) b[a[i]]++;
        S ret = 0;
        rep(i, 1, m + 1) {
            ll di = i;
            S sm = 0;
            while (di <= m) {
                sm += (S)b[di] * di;
                di += i;
            }
            ret += (S)w[i] * sm * sm;
        }
        rep(i, 0, n) ret -= a[i];
        ret /= 2;
        return ret;
    }
    S gcd() {
        // gcd(A, B)
        // O(maxA log maxA)
        T m = *max_element(a.begin(), a.end());
        vector<S> w(m + 1);
        rep(i, 1, m + 1) {
            w[i] += i;
            int di = i + i;
            while (di <= m) {
                w[di] -= w[i];
                di += i;
            }
        }
        vector<ll> b(m + 1);
        rep(i, 0, n) b[a[i]]++;
        S ret = 0;
        rep(i, 1, m + 1) {
            ll di = i;
            S sm = 0;
            while (di <= m) {
                sm += b[di];
                di += i;
            }
            ret += (S)w[i] * sm * sm;
        }
        rep(i, 0, n) ret -= a[i];
        ret /= 2;
        return ret;
    }
};

int main() {
    ll n;
    cin >> n;
    vector<int> a(n);
    rep(i, 0, n) a[i] = i + 1;
    reverse(a.begin(), a.end());
    AllPairOperationSum<int> op(a);
    ll ans = op.floor() + n;
    cout << ans << endl;
}