Case 1 : When N is odd
if ( N % 2 == 1 ) { long long first_maxima = N / 2 ; long long second_maxima = first_maxima + 1 ; long long ans1 = 3e18 , ans2 = 3e18 ; long long from_left = -1 , from_right = -1 ; long long from = -1 ; for ( long long i = 0 ; i < n ; ++ i ) { if ( v [ i ] > first_maxima ) { from = i ; break ; } else { long long diff = first_maxima - v [ i ] ; if ( diff < ans1 ) { ans1 = diff ; from_left = v [ i ] ; } } } from_right = v [ from ] ; long long diff1 = first_maxima - from_left ; long long diff2 = from_right - second_maxima ; if ( diff1 < diff2 ) cout << N << " ▁ " << from_left ; else cout << N << " ▁ " << from_right ; }