what's the closed formed of this sum?

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How can I calculate the closed form of this sum? $$\frac{1}{k} {n\choose k-1} \sum_{j=0}^{n+1-k} (-1)^{j} {n+1-k\choose j} B_{j}.$$ where $B_n$ is the Bernoulli sequence (defined by the exponential generating function$\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$)