#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <bits/stdc++.h>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vmi = vector<mint>;
using vvmi = vector<vmi>;
using vvvmi = vector<vvmi>;
#define all(a) (a).begin(), (a).end()
#define rep2(i, m, n) for (int i = (m); i < (n); ++i)
#define rep(i, n) rep2(i, 0, n)
#define drep2(i, m, n) for (int i = (m)-1; i >= (n); --i)
#define drep(i, n) drep2(i, n, 0)

void solve(){
    
}

class Rational {
public:
    long long num;  
    long long den;  

    Rational(ll numerator = 0, ll denominator = 1) {
        if (denominator == 0) {
            throw std::invalid_argument("denominator must not be zero");
        }
        // 負の符号は分子に寄せる
        if (denominator < 0) {
            numerator = -numerator;
            denominator = -denominator;
        }
        long long g = gcd(numerator, denominator);
        num = numerator / g;
        den = denominator / g;
    }

    
    Rational operator+(const Rational& other) const {
        ll n = num * other.den + den * other.num;
        ll d = den * other.den;
        return Rational(n, d); 
    }
};
ll p(ll a, ll n){
    ll w = 1;
    rep(i, n)w *= a;
    return w;
}

int main(){
    int n; ll a, b; cin >> n >> a >> b;
    vll kaijo(12, 1); rep2(i, 2, 12)kaijo[i] = kaijo[i-1]*(ll)i;
    vll k(n+1); rep(i, n+1)cin >> k[i];
    vll s(n+1);

    Rational ans(0, 1);
    
    rep(i, n+1){
        ll c = p(b, i+1) - p(a, i+1);
        Rational d(k[i]*c, (ll)(i+1));
        ans = ans + d;
    }
    if(ans.num >= 0){
        cout << ans.num / ans.den << endl;
    }else{
        if(ans.num % ans.den == 0){
            cout << ans.num / ans.den << endl;
        }else{
            cout << ans.num / ans.den - 1<< endl;
        }
    }
    


    return 0;
}