// Original C++ solution (slower):
const int LG = 21;

const int FN = 400005;

const long long MOD = 998244353;

const long long INF = 1e9;

const long long INFLL = 1e18;



#include <bits/stdc++.h>



const double PI = acos(-1);



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

typedef vector<int> vi;

typedef vector<ll> vll;

typedef vector<vector<int> > G;



#define forn(i, n) for (int (i) = 0; (i) != (n); (i)++)

#define all(v) (v).begin(), (v).end()

#define rall(v) (v).rbegin(), (v).rend()

#define popcount(x) __builtin_popcount(x)

#define popcountll(x) __builtin_popcountll(x)

#define fi first

#define se second

#define re return

#define pb push_back

#define uniq(x) sort(all(x)); (x).resize(unique(all(x)) - (x).begin())



#ifdef LOCAL

#define dbg(x) cerr << __LINE__ << " " << #x << " " << x << endl

#define ln cerr << __LINE__ << endl

#else

#define dbg(x) void(0)

#define ln void(0)

#endif // LOCAL



int cx[4] = {-1, 0, 1, 0};

int cy[4] = {0, -1, 0, 1};

string Yes[2] = {"No", "Yes"};

string YES[2] = {"NO", "YES"};

string Psb[2] = {"Impossible", "Possible"};

string psb[2] = {"impossible", "possible"};



ll inq(ll x, ll y)

{

    if (!y) re 1 % MOD;

    ll l = inq(x, y / 2);

    if (y % 2) re l * l % MOD * x % MOD;

    re l * l % MOD;

}



ll rev(ll x)

{

    return inq(x, MOD - 2);

}



bool __precomputed_combinatorics = 0;

vector<ll> __fact, __ufact, __rev;



void __precompute_combinatorics()

{

    __precomputed_combinatorics = 1;

    __fact.resize(FN);

    __ufact.resize(FN);

    __rev.resize(FN);

    __rev[1] = 1;

    for (int i = 2; i < FN; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;

    __fact[0] = 1, __ufact[0] = 1;

    for (int i = 1; i < FN; i++) __fact[i] = __fact[i - 1] * i % MOD, __ufact[i] = __ufact[i - 1] * __rev[i] % MOD;

}



ll fact(int x)

{

    if (!__precomputed_combinatorics) __precompute_combinatorics();

    return __fact[x];

}



ll cnk(int n, int k)

{

    if (k < 0 || k > n) return 0;

    if (!__precomputed_combinatorics) __precompute_combinatorics();

    return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;

}



int Root(int x, vector<int> &root)

{

    if (x == root[x]) return x;

    return root[x] = Root(root[x], root);

}



void Merge(int v, int u, vector<int> &root, vector<int> &sz)

{

    v = Root(v, root), u = Root(u, root);

    if (v == u) return;

    if (sz[v] < sz[u])

    {

        sz[u] += sz[v];

        root[v] = u;

    }

    else

    {

        sz[v] += sz[u];

        root[u] = v;

    }

}



int ok(int x, int n)

{

    return 0 <= x && x < n;

}



void bfs(int v, vi &dist, vector<vi> &graph)

{

    fill(all(dist), -1);

    dist[v] = 0;

    vi q = {v};

    for (int i = 0; i < q.size(); i++)

    {

        for (auto u : graph[q[i]])

        {

            if (dist[u] == -1)

            {

                dist[u] = dist[q[i]] + 1;

                q.push_back(u);

            }

        }

    }

}



double sqr(double x)

{

    return x * x;

}



double dist(double x1, double y1, double x2, double y2)

{

    return sqrt(sqr(x1 - x2) + sqr(y1 - y2));

}



signed main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    ll k;

    ll ans = 0;

    cin >> k;

    forn(a, k + 1)

    forn(b, k + 1)

    forn(c, k + 1)

    {

        if (a && b && c) ans += __gcd(__gcd(a, b), c);

    }

    cout << ans;

}



/* Note:

Check constants at the beginning of the code.

    N is set to 4e5 but be careful in problems with large constant factor.

    Setting N in every problem is more effective.

Check corner cases.

    N = 1

No def int long long for now.

Add something here.

*/


// Optimized solution to the same problem (faster):
const int LG = 21;

const int FN = 400005;

const long long MOD = 1e9 + 7;

const long long INF = 1e9;

const long long INFLL = 1e18;



#include <bits/stdc++.h>



using namespace std;



typedef long long ll;

typedef pair<int, int> pii;

typedef pair<ll, ll> pll;

typedef vector<ll> vi;



#define forn(i, n) for (int (i) = 0; (i) != (n); (i)++)

#define all(v) (v).begin(), (v).end()

#define rall(v) (v).rbegin(), (v).rend()

#define popcount(x) __builtin_popcount(x)

#define popcountll(x) __builtin_popcountll(x)

#define fi first

#define se second

#define re return

#define uniq(x) sort(all(x)); (x).resize(unique(all(x)) - (x).begin())



#ifdef LOCAL

#define dbg(x) cerr << __LINE__ << " " << #x << " " << x << endl

#define ln cerr << __LINE__ << endl

#else

#define dbg(x) void(0)

#define ln void(0)

#endif // LOCAL



int cx[4] = {-1, 0, 1, 0};

int cy[4] = {0, -1, 0, 1};



ll inq(ll x, ll y)

{

    if (!y) re 1 % MOD;

    ll l = inq(x, y / 2);

    if (y % 2) re l * l % MOD * x % MOD;

    re l * l % MOD;

}



ll rev(ll x)

{

    return inq(x, MOD - 2);

}



bool __precomputed_combinatorics = 0;

vector<ll> __fact, __ufact, __rev;



void __precompute_combinatorics()

{

    __precomputed_combinatorics = 1;

    __fact.resize(FN);

    __ufact.resize(FN);

    __rev.resize(FN);

    __rev[1] = 1;

    for (int i = 2; i < FN; i++) __rev[i] = MOD - __rev[MOD % i] * (MOD / i) % MOD;

    __fact[0] = 1, __ufact[0] = 1;

    for (int i = 1; i < FN; i++) __fact[i] = __fact[i - 1] * i % MOD, __ufact[i] = __ufact[i - 1] * __rev[i] % MOD;

}



ll fact(int x)

{

    if (!__precomputed_combinatorics) __precompute_combinatorics();

    return __fact[x];

}



ll cnk(int n, int k)

{

    if (k < 0 || k > n) return 0;

    if (!__precomputed_combinatorics) __precompute_combinatorics();

    return __fact[n] * __ufact[n - k] % MOD * __ufact[k] % MOD;

}



signed main()

{

    ios_base::sync_with_stdio(false);

    cin.tie(0);

    cout.tie(0);

    int n = 3, k;

    cin >> k;

    vi c(k + 1);

    ll ans = 0;

    for (int i = 1; i <= k; i++)

    {

        c[i] = inq(k / i, n);

    }

    for (int i = k; i >= 1; i--)

    {

        for (int j = i + i; j <= k; j += i)

        {

            c[i] = (c[i] - c[j] + MOD) % MOD;

        }

    }

    for (int i = 1; i <= k; i++) ans = (ans + c[i] * i) % MOD;

    cout << ans;

}



/* Note:

Check constants at the beginning of the code.

    N is set to 4e5 but be careful in problems with large constant factor.

    Setting N in every problem is more effective.

Check corner cases.

    N = 1

No def int long long for now.

Add something here.

*/


// Fastest known solution for this problem: