#include // {{{ // https://codeforces.com/blog/entry/96344 #pragma GCC optimize("O2,unroll-loops") #pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt") using namespace std; using i32 = int32_t; using u32 = uint32_t; using i64 = int64_t; using u64 = uint64_t; using f64 = double; using f128 = long double; #if __cplusplus >= 202002L template constexpr T MIN = std::numeric_limits::min(); template constexpr T MAX = std::numeric_limits::max(); template [[nodiscard]] static T sc(auto&& x) { return static_cast(x); } template [[nodiscard]] static T sz(auto&& x) { return static_cast(x.size()); } #endif static void NO() { std::cout << "NO\n"; } static void YES() { std::cout << "YES\n"; } template using vec = std::vector; #define all(x) (x).begin(), (x).end() #define rall(x) (x).rbegin(), (x).rend() #define ff first #define ss second #ifdef LOCAL #define db(...) std::print(__VA_ARGS__) #define dbln(...) std::println(__VA_ARGS__) #else #define db(...) #define dbln(...) #endif // }}} struct S{ void method() { } }; void solve() { /* a[i] has factors S, |S| < = log2(MAX_A[i] = 1000) want to find rightmost j > i s.t. no common factors factor -> [a[i] divisible by] iter thru factors of a[i] if i have log2(a[i]) factors; there are 10^6 factors total there are 10 factors here so, factor each a[i], and manually compute */ u32 n; cin >> n; vector a(n); for (auto& e : a) cin >> e; vector> coprime(1001); for (u32 i = 1; i <= 1000; ++i) { for (u32 j = 1; j <= 1000; ++j) { coprime[i][j] = gcd(i, j) == 1; } } vector right(1001, -1); for (u32 i = 0; i < n; ++i) { right[a[i]] = i; } u32 ans = 0; for (i32 i = 1; i <= 1000; ++i) { for (u32 j = 1; j <= 1000; ++j) { if (coprime[i][j] && right[i] != -1 && right[j] != -1) { ans = max(ans, right[i] + right[j] + 2); } } } if (ans == 0) cout << -1; else cout << ans; cout << '\n'; } int main() { // {{{ cin.tie(nullptr)->sync_with_stdio(false); cin.exceptions(cin.failbit); u32 tc = 1; cin >> tc; for (u32 t = 0; t < tc; ++t) { solve(); } return 0; } // }}}