#include <fstream>

using namespace std;

int a, b, c, d, i, s, n;

int main () {
    ifstream fin ("koba.in");
    ofstream fout("koba.out");

    fin>>n>>a>>b>>c;
    a %= 10;
    b %= 10;
    c %= 10;
    if (n == 1) {
        fout<<a;
        return 0;
    }
    if (n == 2) {
        fout<<a+b;
        return 0;
    }
    if (n == 3) {
        fout<<a+b+c;
        return 0;
    }
    s = a + b + c;
    for (i=4;i<=n;i++) {
        d = (c + b*a)%10;
        s += d;

        a = b;
        b = c;
        c = d;
    }
    fout<<s;


/**
    int t[]
    n fiind 10 la a 8-a, un vecor de 100 milioane ar presupune 400 de mega si restrictia din problema
    impune 20 de mega

    fin>>n>>t[1]>>t[2]>>t[3];
    t[1] %= 10;
    t[2] %= 10;
    t[3] %= 10;
    if (n == 1) {
        fout<<t[1];
        return 0;
    }
    if (n == 2) {
        fout<<t[1]+t[2];
        return 0;
    }
    if (n == 3) {
        fout<<t[1]+t[2]+t[3];
        return 0;
    }
    s = t[1] + t[2] + t[3];
    for (i=4;i<=n;i++) {
        t[i] = (t[i-1] + t[i-2]*t[i-3])%10;
        s += t[i];
    }
    fout<<s;
*/
    return 0;
}