#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
using namespace std;
typedef pair<ll, ll> P;
using namespace atcoder;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
#define USE998244353
#ifdef USE998244353
const ll MOD = 998244353;
using mint = modint998244353;
#else
const ll MOD = 1000000007;
using mint = modint1000000007;
#endif

#pragma region //使いがち
const int MAX = 2000001;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 3000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}
ll keta(ll num, ll arity) {
    ll ret = 0;
    while (num) {
        num /= arity;
        ret++;
    }
    return ret;
}
// k進数で見た時のi桁目の数を返す (一番下は0桁目)
ll keta_num(ll num, ll i, ll k) {
    return (num / pow_ll(k, i)) % k;
}
ll ceil(ll n, ll m) {
    // n > 0, m > 0
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
void compress(vector<ll>& v) {
    // [3 5 5 6 1 1 10 1] -> [1 2 2 3 0 0 4 0] 
    vector<ll> u = v;
    sort(u.begin(), u.end());
    u.erase(unique(u.begin(),u.end()),u.end());
    map<ll, ll> mp;
    for (int i = 0; i < u.size(); i++) {
        mp[u[i]] = i;
    }
    for (int i = 0; i < v.size(); i++) {
        v[i] = mp[v[i]];
    }
}

// N以下の素数を全て返す
vector<ll> Eratosthenes( const ll N )
{
    vector<bool> is_prime( N + 1 );
    for( ll i = 0; i <= N; i++ )
    {
        is_prime[ i ] = true;
    }
    vector<ll> P;
    for( ll i = 2; i <= N; i++ )
    {
        if( is_prime[ i ] )
        {
            for( ll j = 2 * i; j <= N; j += i )
            {
                is_prime[ j ] = false;
            }
            P.emplace_back( i );
        }
    }
    return P;
}

vector<pair<ll, ll> > prime_factorize(ll N) {
    vector<pair<ll, ll> > res;
    for (ll a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        ll ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    // 最後に残った数について
    if (N != 1) res.push_back({N, 1});
    return res;
}
#pragma endregion


// 1回しか出てこない素数はどうでもいい
vector<pair<ll, ll> > prime_factorize_here(ll N) {
    vector<pair<ll, ll> > res;
    for (ll a = 2; a * a * a <= N; ++a) {
        if (N % a != 0) continue;
        ll ex = 0; // 指数

        // 割れる限り割り続ける
        while (N % a == 0) {
            ++ex;
            N /= a;
        }

        // その結果を push
        res.push_back({a, ex});
    }

    if (N == 1) return res;

    if (my_sqrt(N) * my_sqrt(N) == N) {
        res.push_back({my_sqrt(N), 2});
    }

    return res;
}

// 0 <= a_1 <= a_2 <= ... <= a_n , a_1 + a_2 + ... + a_n = Sとなる場合のかず
// n, Sは最大で60
// dp[i][j][k] := i番目まで見て、最後の数がjで、和がkになる場合の数
ll solve(ll n, ll s) {
    vector<vector<vector<ll>>> dp(n + 1, vector<vector<ll>>(s + 1, vector<ll>(s + 1, 0)));
    dp[0][0][0] = 1;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= s; j++) {
            for (int k = 0; k <= s; k++) {
                for (int l = max(1, j); l <= s; l++) {
                    int ni = i + 1;
                    int nj = l;
                    int nk = k + l;
                    if (nk > s) continue;
                    dp[ni][nj][nk] += dp[i][j][k];
                }
            }
        }
    }
    ll ret = 0;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= s; j++) {
            ret += dp[i][j][s];
        }
    }
    return ret;
}

int main() {
    ll n;
    cin >> n;
    auto p = prime_factorize(n);
    ll ans = 1;
    for (auto x : p) {
        ans *= solve(60, x.second);
    }
    cout << ans << endl;
    return 0;
}