#include <bits/stdc++.h>

#define show(x) cerr << #x << " = " << x << endl

using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "sz=" << v.size() << "\n[";
    for (const auto& p : v) {
        os << p << ",";
    }
    os << "]\n";
    return os;
}

template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p)
{
    os << "(" << p.first << "," << p.second
       << ")";
    return os;
}


constexpr ll MOD = 1e9 + 7;

template <typename T>
constexpr T INF = numeric_limits<T>::max() / 100;

ll A, N, M;
vector<ll> prime;

template <typename T>
T extgcd(const T a, const T b, T& x, T& y)  // ax+by=gcd(a,b)
{
    T d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1;
        y = 0;
    }
    return d;
}

ll power(const ll n, const ll mod)
{
    if (n == 0) {
        return 1;
    }
    if (n % 2 == 1) {
        return (mod != -1) ? ((power(n - 1, mod) * A) % mod) : (power(n - 1, mod) * A);
    } else {
        const ll pp = power(n / 2, mod);
        return (mod != -1) ? ((pp * pp) % mod) : (pp * pp);
    }
}

template <typename T>
pair<T, T> ChineseRemainderTheorem(const pair<T, T> a1, const pair<T, T> a2)
{
    assert(gcd(a1.first, a2.first) == 1);
    const T p1 = a1.first;
    const T m1 = a1.second;
    const T p2 = a2.first;
    const T m2 = a2.second;
    if (m1 == m2) {
        return make_pair(p1 * p2, m1);
    } else {
        T x, y;
        extgcd(p1, p2, x, y);
        assert(p1 * x + p2 * y == 1);
        x *= (m2 - m1);
        y *= (m2 - m1);
        const T p = p1 * p2;
        return make_pair(p, (((p1 * x + m1) % p) + p) % p);
    }
}

ll over(const ll n, const long double x)  // A↑↑n >= x なら 0, それ以外なら A↑↑n
{
    if (n == 0) {
        return (x <= 1) ? 0 : 1;
    }
    if (x <= 1) {
        return 0;
    }
    const long double lx = log2(x) / log2(A);
    const ll p = over(n - 1, lx);
    if (p == 0) {
        return 0;
    } else {
        return power(p, -1);
    }
}

ll rem(const ll n, const ll mod)  // A↑↑n (mod m)
{
    if (n == 0) {
        return 1 % mod;
    } else if (mod == 1) {
        return 0;
    } else {
        ll num = mod;
        pair<ll, ll> result = make_pair(1, 0);
        for (const ll p : prime) {
            if (num % p == 0) {
                ll beki = 1;
                ll cnt = 0;
                while (num % p == 0) {
                    beki *= p;
                    num /= p;
                    cnt++;
                }
                const ll phi = beki / p * (p - 1);
                pair<ll, ll> fac = make_pair(beki, 1);

                ll a = A;
                int count = 0;
                while (a % p == 0) {
                    count++;
                    a /= p;
                }

                if (count > 0) {
                    const ll ind = over(n - 1, (long double)cnt / count);
                    if (ind == 0) {
                        fac.second = 0;
                    } else {
                        fac.second = power(ind, fac.first);
                    }
                } else {
                    const ll m = rem(n - 1, phi);
                    fac.second = power(m, fac.first);
                }
                result = ChineseRemainderTheorem(result, fac);
            }
            if (num == 1) {
                break;
            }
        }

        return result.second;
    }
}

int main()
{
    cin >> A >> N >> M;
    vector<bool> isprime(M + 1, true);
    for (ll i = 2; i <= M; i++) {
        if (isprime[i]) {
            for (ll j = 2; i * j <= M; j++) {
                isprime[i * j] = false;
            }
        }
    }
    for (ll i = 2; i <= M; i++) {
        if (isprime[i]) {
            prime.push_back(i);
        }
    }

    cout << rem(N, M) << endl;

    return 0;
}