#include <list>

For a sequence X. Xn+1 = ( a * Xn + c ) % p. Now given a, c, X0, n, p, you need to calculate Xn. Input The first line contains an integer T, indicating that there are T test cases (T <= 20); For each case: The only line of the input contains five integers a , c , X0 , n , p. ( a,c,p <= 1e9 , X0,n <= 1e18 ) Output For each test case, output the integer Xn. Sample Input 4 4 3 1 1 7 4 3 7 0 3 4 3 7 1 3 2018 7 15 8 31 Sample Output 0 7 1 10

//由于n=1e18,直接做会超时，用矩阵快速幂，构造：

    a,c     x0,0     x1,0
    0,1  * 1  ,0 =  1  ,0

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
typedef long long ll;
ll b,c,x0,n,p;
struct mat
{
    ll a[2][2];
};
mat mul(mat x,mat y)
{
    mat ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
            {
                ans.a[i][j]=((ans.a[i][j]+((x.a[i][k]%p)*y.a[k][j])%p))%p;
            }
        }
    }
    return ans;
}
ll fastpowerMod(ll n)
{
    mat s;
    memset(s.a,0,sizeof(s.a));
    for(int i=0;i<2;i++)s.a[i][i]=1;
    mat res;
    memset(res.a,0,sizeof(res.a));
    res.a[0][0]=b;
    res.a[0][1]=c;
    res.a[1][0]=0;
    res.a[1][1]=1;
    while(n!=0)
    {
        if(n%2==1)
            s=mul(s,res);
        res=mul(res,res);
        n=n>>1;
    }
    mat result;
    result.a[0][0]=x0;
    result.a[1][0]=1;
    result.a[0][1]=result.a[1][1]=0;
    result=mul(s,result);
    return result.a[0][0];
}
int main()
{
    ll t;
    while(~scanf("%lld",&t))
    {

        while(t--)
        {
            scanf("%lld%lld%lld%lld%lld",&b,&c,&x0,&n,&p);
            if(n==0)
                printf("%lld\n",x0);
            else
                printf("%lld\n",fastpowerMod(n));
        }
    }
    return 0;
}