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#include <fstream>
#define DIM 1000005
using namespace std;
int a[DIM] , b[DIM] , c[DIM] , ans[DIM] , nxt[DIM];
int n , nr;
class InParser {
private:
FILE *fin;
char *buff;
int sp;
char read_ch() {
++sp;
if (sp == 4096) {
sp = 0;
fread(buff, 1, 4096, fin);
}
return buff[sp];
}
public:
InParser(const char* nume) {
fin = fopen(nume, "r");
buff = new char[4096]();
sp = 4095;
}
InParser& operator >> (int &n) {
char c;
while (!isdigit(c = read_ch()) && c != '-');
int sgn = 1;
if (c == '-') {
n = 0;
sgn = -1;
} else {
n = c - '0';
}
while (isdigit(c = read_ch())) {
n = 10 * n + c - '0';
}
n *= sgn;
return *this;
}
} f ("curcubeu.in");
class OutParser {
private:
FILE *fout;
char *buff;
int sp;
void write_ch(char ch) {
if (sp == 50000) {
fwrite(buff, 1, 50000, fout);
sp = 0;
buff[sp++] = ch;
} else {
buff[sp++] = ch;
}
}
public:
OutParser(const char* name) {
fout = fopen(name, "w");
buff = new char[50000]();
sp = 0;
}
~OutParser() {
fwrite(buff, 1, sp, fout);
fclose(fout);
}
OutParser& operator << (int vu32) {
if (vu32 <= 9) {
write_ch(vu32 + '0');
} else {
(*this) << (vu32 / 10);
write_ch(vu32 % 10 + '0');
}
return *this;
}
OutParser& operator << (char ch) {
write_ch(ch);
return *this;
}
OutParser& operator << (const char *ch) {
while (*ch) {
write_ch(*ch);
++ch;
}
return *this;
}
} g ("curcubeu.out");
int find (int pos) {
if (nxt[pos] == 0)
return pos;
int p = pos;
while (nxt[pos]) {
pos = nxt[pos];
}
while (p != pos) {
nxt[p] = pos;
p = nxt[p];
}
return pos;
}
void solve (int left , int right , int val) {
while (left <= right) {
if (ans[left] == 0) {
ans[left] = val;
if (left != right)
nxt[left] = right;
++nr;
if (nr == n - 1) {
return;
}
++left;
}
else {
left = find (left) + 1;
}
}
}
int main() {
f >> n >> a[1] >> b[1] >> c[1];
for (int i = 2; i < n; ++i) {
long long aux = ((long long)a[i - 1] * (long long)i) % n;
a[i] = aux;
aux = ((long long)b[i - 1] * (long long)i) % n;
b[i] = aux;
aux = ((long long)c[i - 1] * (long long)i) % n;
c[i] = aux;
}
for (int i = n - 1; i >= 1; --i) {
if (a[i] > b[i]) {
swap (a[i] , b[i]);
}
solve (a[i] , b[i] , c[i]);
}
for (int i = 1; i < n; ++i) {
g << ans[i] << '\n';
}
return 0;
}