//---------------------------------------------------------------------------
#pragma option -Od
//---------------------------------------------------------------------------
#include <stdio.h>
//---------------------------------------------------------------------------
#define MAX_M   7000
//---------------------------------------------------------------------------
int     T0, T1, a, b, x, y, z, M;
long long       n;
int     cache0[MAX_M];  //=a * T(n-2)^^2 + x * T(n-2)      ; for each T(n-2)
int     cache1[MAX_M];  //=b * T(n-1)^^2 + y * T(n-1) + z  ; for each T(n-1)
//---------------------------------------------------------------------------
void    init()
{
        FILE    *F= fopen("rsir.in", "rt");
        fscanf(F, "%d %d %d %d %d %d %d %d %Ld",
                &T0, &T1, &a, &b, &x, &y, &z, &M, &n);
        fclose(F);
        T0%= M;
        T1%= M;
        for (int I= M; --I>= 0; ) {
                int     sqr= I* I % M;
                cache0[I]= (a* sqr+ x* I   ) % M;
                cache1[I]= (b* sqr+ y* I+ z) % M;
        }
}
//---------------------------------------------------------------------------
#define NEXT1(lo, hi)                                   \
        {                                               \
                int     sav= hi;                        \
                hi= (cache0[lo]+ cache1[hi]) % M;       \
                lo= sav;                                \
        }       //one iteration...
//---------------------------------------------------------------------------
#define NEXT2(lo, hi)                                   \
        {                                               \
                lo= (cache0[lo]+ cache1[hi]) % M;       \
                hi= (cache0[hi]+ cache1[lo]) % M;       \
        }       //two iterations...
//---------------------------------------------------------------------------
int     calc()
{
        int     p0, p1, q0, q1;
        p0= q0= T0;
        p1= q1= T1;
        int     index= 0;       //index of (p0, p1)
        do {
                NEXT1(p0, p1);
                index++;
                NEXT2(q0, q1);
        } while (p0!= q0 || p1!= q1);
        int     s0= p0, s1= p1;
        //acum perechea (p0, p1) face parte din ciclu, la index-ul index;
        //aplicam acelasi algoritm pentru a afla lungimea ciclului...
        int     cycle= 0;
        do {
                NEXT1(p0, p1);
                cycle++;
                NEXT2(q0, q1);
        } while (p0!= q0 || p1!= q1);

        n= (n- index) % cycle + index;  //n se reduce la un singur ciclu...
        int     r0= T0, r1= T1;
        if (n>= index) {
                n-= index;
                r0= s0;
                r1= s1; //evita primele calcule...
        }
        for (int I= ((int)n)>> 1; --I>= 0; )
                NEXT2(r0, r1);
        if (((int)n) & 1)
                return  r1;
        return  r0;
}
//---------------------------------------------------------------------------
void    outp()
{
        FILE    *G= fopen("rsir.out", "wt");
        fprintf(G, "%d\n", calc());
        fclose(G);
}
//---------------------------------------------------------------------------
int     main(int argc, char* argv[])
{
        init();
        outp();
        return  0;
}
//---------------------------------------------------------------------------
/*
Rsir

Construim un sir recurent astfel:

Tn = a * T(n-2)^^2 + b * T(n-1)^^2 + x * T(n-2) + y * T(n-1) + z
Cerinta

Fiind date T0, T1, a, b, x, y, z si n calculati Tn modulo un numar natural M.
Date de intrare

Fisierul de intrare rsir.in contine pe prima linie numerele naturale T0, T1, a, b, x, y, z, M si n, separate prin spatiu, cu semnificatia din enunt.
Date de iesire

Fisierul de iesire rsir.out va contine o singura linie pe care va fi scris un numar natural reprezentand Tn modulo M.
Restrictii
0 <= a, b, x, y, z <= 1.000
0 <= T0, T1 <= 1.000.000.000
0 <= n <= 1016
1 <= M <= 7.000
Exemplursir.in	                rsir.out
1 1 0 0 1 1 0 1000 7	        21

Explicatie

Termenii sirului sunt:
T0=1
T1=1
T2=0*12+0*12+1*1+1*1+0=2
T3=0*12+0*22+1*1+1*2+0=3
T4=0*22+0*32+1*2+1*3+0=5
T5=0*32+0*52+1*3+1*5+0=8
T6=0*52+0*82+1*5+1*8+0=13
T7=0*82+0*132+1*8+1*13+0=21
Rezultatul este T7 mod 1000 = 21.
*/
//---------------------------------------------------------------------------