#include <iostream>
#include<cstring>
#include <cmath>
#include <map>
#include <cstdio>
using namespace std;
const int maxn = 1e5+10;
typedef long long ll;
ll ex_gcd(ll a, ll b, ll &x, ll &y){
    if(a == 0&& b == 0) return -1;
    if(b == 0){
        x = 1, y = 0;
        return a;
    }
    ll d = ex_gcd(b, a%b, y, x);
    y -= a/b*x;
    return d;
}
ll inv(ll a, ll n){
    ll x, y;
    ll d = ex_gcd(a, n, x, y);
    if(d == -1) return -1;
    else return (x+n)%n;
}
ll pow_mod(ll a, ll b, ll p){
    ll ans = 1;
    while(b){
        if(b&1){
            ans = ans*a%p;
        }
        a = a*a%p;
        b>>=1;
    }
    return ans;
}
ll mul_mod(ll a, ll b, ll p){
    return (a%p)*(b%p)%p;
}
/*
return x, a^x\equav b(mod n), and n is a prime.
*/
ll bsgs(ll a, ll b, ll n){
    ll m, v, e=1, i;
    m = sqrt(n+0.5);
    //cout<<pow_mod(a, m, n)<<endl;
    v = inv(pow_mod(a, m, n), n);
    map<int, int> x;
    x.clear();
    x[1] = 0;
    //cout<<v<<endl;
    for(int i=1; i<m; i++){
        e = mul_mod(e, a, n);
        //cout<<e<<endl;
        if(!x.count(e)) x[e] = i;
    }
    for(int i=0; i<m ;i++){
        if(x.count(b)) return i*m+x[b];
        b = mul_mod(b, v, n);
    }
    return -1;
}
int main()
{
    ll b, n, p;
    while(~scanf("%lld%lld%lld", &p, &b, &n)){
        ll ans = bsgs(b, n, p);
        if(ans == -1) printf("no solution\n");
        else printf("%lld\n", ans);
    }
    return 0;
}