For an integer $n$, we define the square prime factors of $n$ to be the primes whose square divides $n$. For example, the square prime factors of $1500=2^2\times 3\times 5^3$ are $2$ and $5$.

Let $C_k(N)$ be the number of integers between $1$ and $N$ inclusive with exactly $k$ square prime factors. It can be shown that with growing $N$ the ratio $\frac{C_k(N)}{N}$ gets arbitrarily close to a constant $c_k^{\mathrm{\infty }}$, as suggested by the table below.

$$\begin{array}{|cccccc|}\hline & k=0& k=1& k=2& k=3& k=4\\ C_k(10)& 7& 3& 0& 0& 0\\ C_k({10}^2)& 61& 36& 3& 0& 0\\ C_k({10}^3)& 608& 343& 48& 1& 0\\ C_k({10}^4)& 6083& 3363& 533& 21& 0\\ C_k({10}^5)& 60794& 33562& 5345& 297& 2\\ C_k({10}^6)& 607926& 335438& 53358& 3218& 60\\ C_k({10}^7)& 6079291& 3353956& 533140& 32777& 834\\ C_k({10}^8)& 60792694& 33539196& 5329747& 329028& 9257\\ C_k({10}^9)& 607927124& 335389706& 53294365& 3291791& 95821\\ c_k^{\mathrm{\infty }}& \frac{6}{{\pi }^2}& 3.3539\times {10}^{-1}& 5.3293\times {10}^{-2}& 3.2921\times {10}^{-3}& 9.7046\times {10}^{-5}\\ \hline\end{array}$$ Find $c_7^{\mathrm{\infty }}$. Give the result in scientific notation rounded to $5$ significant digits, using a $e$ to separate mantissa and exponent. E.g. if the answer were $0.000123456789$, then the answer format would be $1.2346\mathrm{e}-4$.