Magic Sequence2

/// --- 31623 = O(181960626) --- ///
/// --- Best Base is: Base31623 for (2 digit for 1e9) --- ///

#include <bits/stdc++.h>
using namespace std;

const int N = 1e6 + 6e1;
const int M = 31623;

int n ,cnt1[M] ,cnt2[M];
int s1[N][3] ,s2[N][3] ,R[N];
int a ,b ,c ,x ,y;
bool vs[N];

int main()
{
    int Tc;
    scanf("%d" ,&Tc);

    while( Tc-- )
    {
        scanf("%d%d%d%d%d%d" ,&n ,&a ,&b ,&c ,&x ,&y);

        s1[0][0] = a;
        s1[0][1] = a % M;
        s1[0][2] = a / M;

        int L1 = s1[0][1] ,R1 = s1[0][1];
        int L2 = s1[0][2] ,R2 = s1[0][2];
        for(int i = 1 ; i < n ; i++)
        {
            s1[i][0] = ((1ll * s1[i - 1][0] % c * b % c) % c + a % c) % c;
            s1[i][1] = s1[i][0] % M;
            s1[i][2] = s1[i][0] / M;

            L1 = min(L1 ,s1[i][1]);
            R1 = max(R1 ,s1[i][1]);
            L2 = min(L2 ,s1[i][2]);
            R2 = max(R2 ,s1[i][2]);
        }

        int K1 = R1 - L1 + 1;
        int K2 = R2 - L2 + 1;
        int K = max(K1 , K2);

        /// ----------------------------------------------- ///

        for(int i = 0 ; i < n ; i++)
        {
            cnt1[ s1[i][1] - L1 ] ++;
            cnt2[ s1[i][2] - L2 ] ++;
        }

        for(int i = 1 ; i < K ; i++)
        {
            if(i < K1) cnt1[i] += cnt1[i - 1];
            if(i < K2) cnt2[i] += cnt2[i - 1];
        }

        int j = n - 1;
        for(int i = n - 1 ; i >= 0 ; i--)
        {
            s2[ cnt1[ s1[i][1] - L1 ] - 1 ][0] = s1[i][0];
            s2[ cnt1[ s1[i][1] - L1 ] - 1 ][1] = s1[i][1];
            s2[ cnt1[ s1[i][1] - L1 ] - 1 ][2] = s1[i][2];
            vs[ cnt1[ s1[i][1] - L1 ] - 1 ] = 1;
            cnt1[ s1[i][1] - L1 ] --;

            while( vs[j] )
            {
                R[ cnt2[ s2[j][2] - L2 ] - 1 ] = s2[j][0];
                cnt2[ s2[j][2] - L2 ] --;
                j--;
            }
        }

        /// ----------------------------------------------- ///

        int v = 0;
        for(int i = 0 ; i < max(n ,K) ; i++)
        {
            if( i < n )
            {
                v = ((1ll * v % y * x % y) % y + R[i] % y) % y;
                vs[i] = 0;
            }

            if( i < K1 ) cnt1[i] = 0;
            if( i < K2 ) cnt2[i] = 0;
        }

        printf("%d\n" ,v);
    }

    return 0;
}