$$\begin{gathered} \prod _{d=1}^n{d}^{\sum [\gcd (\frac{s_1}{d},\frac{s_2}{d},\dots ,\frac{s_k}{d})=1]} \\ \prod _{d=1}^n{d}^{\sum _{i=1}^{\frac{n}{d}}\mu (i)C_{f[id]}^k} \end{gathered}$$
筛出质数，然后得到$phi(mod)$，即可欧拉降幂写（提供一个思路整体复杂度$n{\log }^{2}n$了）。

#pragma GCC optimize(3,"Ofast","inline")
#include <bits/stdc++.h>using namespace std;const int M = 1e7 + 10, N = 8e4 + 10, maxn = 80000;int prime[M], mu[M], cnt;int f[N], n, k;long long C[N][32], mod, phi;bool st[M];inline long long read() {long long x = 0;char c = getchar();while (c < '0' || c > '9') {c = getchar();}while (c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return x;
}void Init() {mu[1] = 1;for (int i = 2; i < M; i++) {if (!st[i]) {prime[++cnt] = i;mu[i] = -1;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < M; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {break;}mu[i * prime[j]] = -mu[i];}}
}long long Phi(long long n) {long long ans = n;for (int i = 1; i <= cnt && 1ll * prime[i] * prime[i] <= n; i++) {if (n % prime[i] == 0) {ans = ans / prime[i] * (prime[i] - 1);while (n % prime[i] == 0) {n /= prime[i];}}}if (n > 1) {ans = ans / n * (n - 1);}return ans;
}long long mul(long long x, long long y) { return (__int128)x * y % mod; 
}long long quick_pow(long long a, long long n) {long long ans = 1;while (n) {if (n & 1) {ans = mul(ans, a);}a = mul(a, a);n >>= 1;}return ans;
}int main() {// freopen("data05.in", "r", stdin);// freopen("out.txt", "w", stdout);Init();int T;T = read();while (T--) {n = read(), k = read(), mod = read();for (int i = 1; i < N; i++) {f[i] = 0;}for (int i = 1, x; i <= n; i++) {x = read();f[x]++;}for (int i = 1; i <= maxn; i++) {for (int j = 2 * i; j <= maxn; j += i) {f[i] += f[j];}}phi = Phi(mod);C[0][0] = 1;for (int i = 1; i <= maxn; i++) {C[i][0] = 1;for (int j = 1; j <= 30; j++) {C[i][j] = C[i - 1][j - 1] + C[i - 1][j];if (C[i][j] > phi) {C[i][j] -= phi;}}}long long ans = 1;for (int d = 2; d <= maxn; d++) {long long cur = 0;for (int i = 1; i <= maxn / d; i++) {if (mu[i] == -1) {cur -= C[f[i * d]][k];if (cur < 0) {cur += phi;}}else if (mu[i] == 1) {cur += C[f[i * d]][k];if (cur > phi) {cur -= phi;}}}ans = mul(ans, quick_pow(d, cur));}printf("%lld\n", ans);}return 0;
}