LL dp[301][2501];
dp[0][1] = 1;
for (int i = 0; i < n; i++) {
    for (int j = 0; j < 2500; j++) {
        dp[i + 1][j * a[i] % (p * q)] += dp[i][j];
    }
}

LL answer = dp[n][0];