Finite Galois module whose Ш¹ is nonzero?

In algebraic number theory, we constantly make use of the nine-term Poitou-Tate sequence: Let $$K$$ be a number field and $$M$$ a finite $$K$$-Galois module. Then we have the nine-term exact sequence $$H^0(K, M) \to \prod' H^0(K_v,M) \to H^2(K, M^\vee)^\vee \mathop{\to}\limits^\delta H^1(K, M) \mathop{\to}^{\operatorname{loc}} \prod' H^1(K_v,M) \to \cdots$$ The kernel of the localization map $$\operatorname{loc}$$, which is also the image of the connecting map $$\delta$$, is traditionally called $$Ш^1(K, M)$$. It is frequently remarked (e.g. in Neukirch, Schmidt, and Wingberg's Cohomology of Number Fields, Definition 8.6.2), that $$Ш^1$$ is finite but not necessarily $$0$$. However, in all the examples I've been able to compute explicitly (e.g. $$M$$ cyclic of prime order), $$Ш^1 = 0$$. Is there a ready example of a number field $$K$$ and a finite module $$M$$ such that $$Ш^1(K, M) \neq 0$$?

Wang's conterexample to Grunwald's theorem: $$K=\mathbb{Q}(\sqrt{7})$$ and $$M=\mu_8$$. Then $$H^1(K,M) \cong K^\times/(K^\times)^8$$. Now $$16$$ is not an $$8$$-th power in this field but locally an $$8$$-th power everywhere. Your group is cyclic of order $$2$$. See wikipedia.