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We prove that for $0\\le i\\le k-1$, \\[ b_i(S) \\le{1/2}(\\sum_{j=0}^{min\\{s,k-i\\}}{{s}\\choose j}{{k+1}\\choose {j}}2^{j}). \\] In particular, for $2\\le s\\le \\frac{k}{2}$, we have \\[ b_i(S)\\le {1/2} 3^{s}{{k+1}\\choose {s}} \\leq {1/2} (\\frac{3e(k+1)}{s})^s. \\] This improves the bound of $k^{O(s)}$ proved by Barvinok. 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