#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <fstream>
#include <vector>
#include <string>
#include <math.h>
#include <cmath>
#include <set>
#include <map>
#include <cassert>
#include <algorithm>
#include <ctime>
#include <iomanip>
#include <unordered_map>
//#include <thread>

using namespace std;

#define mp make_pair
#define sz size()
#define all(x) x.begin(),x.end()
#define forn(i,n) for (int i = 0; i<int(n); i++)
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
const ll INF = 1e15 + 7;
const ld EPS = 1e-14;

ll diskret_logarifm(ll a, ll b, ll m){
    ll n = (ll)sqrt((ld)m) + 1; // a^(p*n-q) = b(mod m) => a^(p*n) = b*a^q(mod m)  p - переменная

    ll an = 1; // a^n mod m
    for (int i = 0; i < n; i++)
        an = (an*a) % m;

    unordered_map<int, int> vals;
    for (ll i = 1, cur = an; i <= n; i++){
        if (!vals.count(cur))
            vals[cur] = i;
        cur = (cur*an) % m;
    }

    for (ll i = 0, cur = b; i <= n; i++){
        if (vals.count(cur)){
            ll ans = vals[cur] * n - i; // p*n-q
            if (ans < m)
                return ans;
        }
        cur = (cur*a) % m;
    }

    return -1;
}

void solve_interval(ll p, ll a, ll l, ll r){
    vector<int> ans;
    for (int cur = l; cur <= r; cur++){
        if (diskret_logarifm(a, cur, p) != -1)
            ans.push_back(cur);
    }

    for (int i = 0; i < ans.size(); i++)
        printf("%d ", ans[i]);
}

void solve_powers(ll p, ll a, ll l, ll r){
    unordered_map<int, bool> M;
    vector<int> ans;

    ll cur = 1;
    int cnter = 0;
    while (!M.count(cur)){
        cnter++;
        M[cur] = true;
        if (cur >= l && cur <= r)
            ans.push_back(cur);
        cur = (cur*a) % p;
    }

    sort(all(ans));
    for (int i = 0; i < ans.size(); i++)
        printf("%d ", ans[i]);
}

//#define DEBUG
int main(){
#ifdef DEBUG
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#else
    freopen("list-powers.in", "r", stdin);
    freopen("list-powers.out", "w", stdout);
#endif

    ll p, a, l, r;
    cin >> p >> a >> l >> r;
    if (r - l <= 1001){
        solve_interval(p, a, l, r);
    }
    else{
        solve_powers(p, a, l, r);
    }

#ifdef DEBUG
    cerr << "Time elapsed: " << (ld)clock() / CLOCKS_PER_SEC << " sec." << endl;
#endif
    return 0;
}