If \( u(x,t) = A e^{-t} \sin x \) solves the following initial boundary value problem:
\[ \frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2}, 0 < x < \pi, t > 0, \] \[ u(0,t) = u(\pi,t) = 0, t > 0, \] \[ u(x,0) = \begin{cases} 60, & 0 < x \leq \frac{\pi}{2}, \\ 40, & \frac{\pi}{2} < x < \pi, \end{cases} \] then \( \pi A = \underline{\hspace{1cm}} \).