#define _USE_MATH_DEFINES
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>
#include<unordered_set>
#include<random>
#include <array>
#include <complex>
#include<chrono>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
#define LP(I,S,G) for (long long int I = (S); I < (G); ++I)
#define IN(X) 	for (int in = 0; in < X.size(); in++)cin >> X[in]
#define OUT(X) 	for (int in = 0; in < X.size(); in++)cout << X[in]<<" "
#define SORT(X) sort((X).begin(), (X).end())
#define CSORT(X,Y) sort(X.begin(), X.end(),Y)
#define COPY(X,Y) copy(X.begin(), X.end(), Y.begin())
#define ALL(X,Y) for (auto &(X) :(Y))
#define FULL(a)  (a).begin(),(a).end()
#define BFS(Q,S) for(Q.push(S);Q.size()!=0;Q.pop())
typedef long long int ll;
typedef unsigned long long int ull;
long long int M = 998244353;

chrono::system_clock::time_point starttime;
using namespace std::chrono;

inline float getTime() {
	return duration_cast<milliseconds>(system_clock::now() - starttime).count();
}


int dx[] = { -1,0,1,0 }, dy[] = { 0,1,0,-1 };

ll MAX(ll A, ll B) { return ((A) > (B) ? (A) : (B)); }
ll MIN(ll A, ll B) { return ((A) < (B) ? (A) : (B)); }
inline long long int xor128() {
	static long long int x = 123456789, y = 362436069, z = 521288629, w = 88675123;
	long long int t = (x ^ (x << 11));
	x = y; y = z; z = w;
	return (w = (w ^ (w >> 19)) ^ (t ^ (t >> 8)));
}

struct HashPair {

	//注意 constがいる
	template<class T1, class T2>
	size_t operator()(const pair<T1, T2> &p) const {

		//first分をハッシュ化する
		auto hash1 = hash<T1>{}(p.first);

		//second分をハッシュ化する
		auto hash2 = hash<T2>{}(p.second);

		//重複しないようにハッシュ処理
		size_t seed = 0;
		seed ^= hash1 + 0x9e3779b9 + (seed << 6) + (seed >> 2);
		seed ^= hash2 + 0x9e3779b9 + (seed << 6) + (seed >> 2);
		return seed;
	}
};



int main() {
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	vector<ll> dp(n * 2, a*n);
	dp[1] = 0;
	ll kai = 1, border = a * n;
	ll ku = 0;
	LP(i, 1, 2*n) {
		kai *= i;
		if (kai >= n)ku = 1;
		kai %= n;
		dp[(i + 1) % n] = min(dp[(i + 1) % n], dp[i] + a);
		dp[kai + ku * n] = min(dp[kai + ku * n], dp[i] + c);
		ll p = i, cost = b, pu = 0;
		if (p >= n)pu = 1;
		p %= n;
		while (cost <= border) {
			dp[p + pu * n] = min(dp[p + pu * n], dp[i] + cost);
			p *= i;
			if (p >= n)pu = 1;
			p %= n;
			cost *= b;
		}
	}
	ll fc = a*n;
	LP(i, n + 1, 2 * n)fc = min(fc, dp[i]);
	cout << min(min(dp[0], dp[n]), fc + c);
	return 0;
}