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inFullMobile Language Wars: Round 1 - JavaScript Solution 2 by @rsp - See: https://gist.github.com/rsp/d8bdbafa09f24f99eebc8ed60fe205c8
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| const e = a => b => c => c ? e(b(a))(b)(c - 1) : a; | |
| const f = a => b => c => e(c)(b)(a(a => a + 1)(0) - 1); |
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