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#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <iostream>
#include <fstream>
//#include <string.h>
#define KMAX 3
#define MOD 666013
using namespace std;
void multiply_matrix(int A[KMAX][KMAX], int B[KMAX][KMAX], int C[KMAX][KMAX]) {
int tmp[KMAX][KMAX];
// tmp = A * B
for (int i = 0; i < KMAX; ++i) {
for (int j = 0; j < KMAX; ++j) {
unsigned long long sum = 0; // presupun ca suma incape pe 64 de biti
for (int k = 0; k < KMAX; ++k) {
sum += 1LL * A[i][k] * B[k][j];
}
tmp[i][j] = sum % MOD;
}
}
// C = tmp
for(int i=0;i<KMAX;i++)
for(int j=0;j<KMAX;j++)
C[i][j] = tmp[i][j];
}
void power_matrix(int C[KMAX][KMAX], int p, int R[KMAX][KMAX]) {
// tmp = I (matricea identitate)
int tmp[KMAX][KMAX];
for (int i = 0; i < KMAX; ++i) {
for (int j = 0; j < KMAX; ++j) {
tmp[i][j] = (i == j) ? 1 : 0;
}
}
while (p != 1) {
if (p % 2 == 0) {
multiply_matrix(C, C, C); // C = C*C
p /= 2; // ramane de calculat C^(p/2)
} else {
// reduc la cazul anterior:
multiply_matrix(tmp, C, tmp); // tmp = tmp*C
--p; // ramane de calculat C^(p-1)
}
}
// avem o parte din rezultat in C si o parte in tmp
multiply_matrix(C, tmp, R); // rezultat = tmp * C
}
int iepuri(int x,int y,int z,int a,int b,int c,int n){
if(n==0) return x;
if(n==1) return y;
if(n==2) return z;
int matrix[3][3] = {{0,1,0},{0,0,1},{c,b,a}};
int rezultat[3][3];
power_matrix(matrix,n-2,rezultat);
return (1LL*rezultat[2][0]*x%MOD + 1LL*rezultat[2][1]*y%MOD + 1LL*rezultat[2][2]*z%MOD)%MOD;
}
int main(){
ifstream in("iepuri.in");
ofstream out("iepuri.out");
int nr_linii,x,y,z,a,b,c,n;
in>>nr_linii;
for(int i=0;i<nr_linii;i++){
in>>x>>y>>z>>a>>b>>c>>n;
out<<iepuri(x,y,z,a,b,c,n);
if(i!=nr_linii-1)
out<<endl;
}
}