以下除法均为向下取整， 定义$f(a,b,c,n,k_1,k_2)=\sum _{x=0}^n{x}^{k_1}{\left(\frac{a\times x+b}{c}\right)}^{k_2}$。
$$\sum _{x=0}^n{x}^{k_1}{\left(\frac{a\times x+b}{c}\right)}^{k_2},(1\le n,a,c\le 10^9,0\le b\le 10^9,0\le k_1+k_2\le 10)$$
$a\ge c,a^{\prime }=a\%c,k=\frac{a}{c}$：
$$\begin{gathered} \sum _{x=0}^n{x}^{k_1}{\left(kx+\frac{a^{\prime }\times x+b}{c}\right)}^{k_2} \\ \sum _{x=0}^n{x}^{k_1}\sum _{i=0}^{k_2}{C}_{k_2}^i(kx{)}^i{\left(\frac{a^{\prime }\times x+b}{c}\right)}^{k_2-i} \\ \sum _{i=0}^{k_2}\left(C_{k_2}^i{k}^i\right)\sum _{x=0}^n{x}^{i+k_1}{\left(\frac{a^{\prime }\times x+b}{c}\right)}^{k_2-i} \\ f(a,b,c,n,k_1,k_2)=\sum _{i=0}^{k_2}{C}_{k_2}^i{k}^{i}f(a^{\prime },b,c,n,i+k_1,k_2-i) \end{gathered}$$
$b\ge c,b^{\prime }=b\%c,k=\frac{b}{c}$：
$$\begin{gathered} \sum _{x=0}^n{x}^{k_1}{\left(\frac{a\times x+b^{\prime }}{c}+k\right)}^{k_2} \\ \sum _{x=0}^n{x}^{k_1}\sum _{i=0}^{k_2}{C}_{k_2}^i{k}^i{\left(\frac{a\times x+b^{\prime }}{c}\right)}^{k_2-i} \\ \sum _{i=0}^{k_2}{C}_{k_2}^i{k}^i\sum _{x=0}^n{x}^{k_1}{\left(\frac{a\times x+b^{\prime }}{c}\right)}^{k_2-i} \\ f(a,b,c,n,k_1,k_2)=\sum _{i=0}^{k_2}{C}_{k_2}^i{k}^{i}f(a,b^{\prime },c,n,k_1,k_2-i) \end{gathered}$$
$a<c,b<c,m=\frac{a\times n+b}{c}$：
$$\begin{gathered} n^k=\sum _{i=1}^n{i}^k-(i-1{)}^k \\ \sum _{x=0}^n{x}^{k_1}{\left(\frac{a\times x+b}{c}\right)}^{k_2} \\ \sum _{x=0}^n{x}^{k_1}\sum _{i=1}^m\left(i^{k_2}-(i-1{)}^{k_2}\right)[\frac{a\times x+b}{c}\ge i] \\ \sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^n{x}^{k_1}[\frac{a\times x+b}{c}\ge i+1] \\ \sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^n{x}^{k_1}-\sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^{\frac{c\times i+c-b-1}{a}}{x}^{k_1} \end{gathered}$$
前项可以通过插值得到，我们考虑后项求和，

不难得到$(i+1{)}^{k_2}-i^{k_2}$，是一个$k_2-1$次的多项式，我们可以通过插值预处理出其系数，设其为$A(x)$。

对于$\sum _{i=0}^{\frac{c\times i+c-b-1}{a}}{x}^{k_1}$，也是一个多项式，最高次幂为$k_1+1$，设其为$B(x)$。
$$\begin{gathered} \sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^n{x}^{k_1}-\sum _{x=0}^{m-1}\sum _{i=0}^{k_2-1}{A}_i{x}^i\sum _{j=0}^{k_1+1}{B}_j{\left(\frac{c\times x+c-b-1}{a}\right)}^j \\ \sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^n{x}^{k_1}-\sum _{i=0}^{k_2-1}{A}_i\sum _{j=0}^{k_1+1}{B}_j\sum _{x=0}^{m-1}{x}^i{\left(\frac{c\times x+c-b-1}{a}\right)}^j \\ f(a,b,c,n,k_1,k_2)=\sum _{i=0}^{m-1}\left((i+1{)}^{k_2}-i^{k_2}\right)\sum _{x=0}^n{x}^{k_1}-\sum _{i=0}^{k_2-1}\sum _{j=0}^{k_1+1}{A}_i\times B_j\times f(a^{\prime },b^{\prime },c^{\prime },m-1,i,j) \end{gathered}$$
只要预处理各种插值系数，即可在$O(T\times k^4\log n)$的复杂度内求解。

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 15;
const int P = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {x = 0; int f = 1;char c = getchar();for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';x *= f;
}
template <typename T> void write(T x) {if (x < 0) x = -x, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {write(x);puts("");
}
struct info {int a[MAXN][MAXN];
};
int sum[MAXN][MAXN];
int binom[MAXN][MAXN];
int power(int x, int y) {if (y == 0) return 1;int tmp = power(x, y / 2);if (y % 2 == 0) return 1ll * tmp * tmp % P;else return 1ll * tmp * tmp % P * x % P;
}
void update(int &x, int y) {x += y;if (x >= P) x -= P;
}
info func(int n, int a, int b, int c) {assert(n >= 0 && a >= 0 && b >= 0 && c >= 0);info ans;memset(ans.a, 0, sizeof(ans.a));if (a == 0 || (1ll * a * n + b) / c == 0) {for (int k1 = 0; k1 <= 10; k1++) {int mul = 0, now = 1;for (int i = 0; i <= k1 + 1; i++) {update(mul, 1ll * now * sum[k1][i] % P);now = 1ll * now * n % P;}int base = (1ll * a * n + b) / c % P; now = 1;for (int k2 = 0; k1 + k2 <= 10; k2++) {ans.a[k1][k2] = 1ll * now * mul % P;now = 1ll * now * base % P;}}return ans;}if (a >= c) {info tmp = func(n, a % c, b, c);for (int k1 = 0; k1 <= 10; k1++)for (int k2 = 0; k1 + k2 <= 10; k2++) {int now = 1, base = a / c;for (int i = 0; i <= k2; i++) {update(ans.a[k1][k2], 1ll * binom[k2][i] * now % P * tmp.a[k1 + i][k2 - i] % P);now = 1ll * now * base % P;}}return ans;}if (b >= c) {info tmp = func(n, a, b % c, c);for (int k1 = 0; k1 <= 10; k1++)for (int k2 = 0; k1 + k2 <= 10; k2++) {int now = 1, base = b / c;for (int i = 0; i <= k2; i++) {update(ans.a[k1][k2], 1ll * binom[k2][i] * now % P * tmp.a[k1][k2 - i] % P);now = 1ll * now * base % P;}}return ans;}int m = (1ll * a * n + b) / c;info tmp = func(m - 1, c, c - b - 1, a);for (int k1 = 0; k1 <= 10; k1++) {int all = 0, now = 1;for (int i = 0; i <= k1 + 1; i++) {update(all, 1ll * now * sum[k1][i] % P);now = 1ll * now * n % P;}for (int k2 = 0; k1 + k2 <= 10; k2++) {ans.a[k1][k2] = 1ll * power(m, k2) * all % P;for (int i = 0; i <= k2 - 1; i++)for (int j = 0; j <= k1 + 1; j++) {int coef = 1ll * binom[k2][i] * sum[k1][j] % P;update(ans.a[k1][k2], P - 1ll * coef * tmp.a[i][j] % P);}}}return ans;
}
void Lagrange(int n, int *x, int *y, int *a) {static int p[MAXN], q[MAXN];memset(p, 0, sizeof(p)); p[0] = 1;for (int i = 1; i <= n; i++) {for (int j = i - 1; j >= 0; j--) {p[j + 1] = (p[j + 1] + p[j]) % P;p[j] = (P - 1ll * p[j] * x[i] % P) % P;}}for (int i = 1; i <= n; i++) {memset(q, 0, sizeof(q));for (int j = n - 1; j >= 0; j--)q[j] = (p[j + 1] + 1ll * q[j + 1] * x[i]) % P;int now = 1;for (int j = 1; j <= n; j++)if (j != i) now = 1ll * now * (x[i] - x[j]) % P;now = power((P + now) % P, P - 2);for (int j = 0; j <= n; j++)q[j] = 1ll * q[j] * now % P;for (int j = 0; j <= n; j++)a[j] = (a[j] + 1ll * q[j] * y[i]) % P;}
}
int main() {for (int i = 0; i <= 10; i++) {static int pos[MAXN], now[MAXN];now[0] = power(0, i);for (int j = 1; j <= i + 2; j++) {now[j] = (now[j - 1] + power(j, i)) % P;pos[j] = j;}Lagrange(i + 2, pos, now, sum[i]);}for (int i = 0; i <= 10; i++) {binom[i][0] = 1;for (int j = 1; j <= i; j++)binom[i][j] = binom[i - 1][j - 1] + binom[i - 1][j];}int T; read(T);while (T--) {int n, a, b, c, k1, k2;read(n), read(a), read(b), read(c), read(k1), read(k2);writeln(func(n, a, b, c).a[k1][k2]);}return 0;
}