text: Find sum of xor of all unordered triplets of the array | Iterative Function to calculate ( x ^ y ) % p in O ( log y ) ; Initialize result ; Update x if it is more than or equal to p ; If y is odd , multiply x with result ; y must be even now y = y >> 1 ; y = y / 2 ; Returns n ^ ( - 1 ) mod p ; Returns nCr % p using Fermat 's little theorem. ; Base case ; Fill factorial array so that we can find all factorial of r , n and n - r ; Function returns sum of xor of all unordered triplets of the array ; Iterating over the bits ; Number of elements whith k 'th bit  1 and 0 respectively ; Checking if k 'th bit is 1 ; Adding this bit 's part to the answer ; Driver Code
code: function power ( x , y , p ) { let res = 1 ; x = x % p ; while ( y > 0 ) { if ( ( y & 1 ) == 1 ) res = ( res * x ) % p ; x = ( x * x ) % p ; } return res ; } function modInverse ( n , p ) { return power ( n , p - 2 , p ) ; } function nCrModPFermat ( n , r , p ) { if ( r == 0 ) return 1 ; if ( n < r ) return 0 ; let fac = Array . from ( { length : n + 1 } , ( _ , i ) => 0 ) ; fac [ 0 ] = 1 ; for ( let i = 1 ; i <= n ; i ++ ) fac [ i ] = fac [ i - 1 ] * i % p ; return ( fac [ n ] * modInverse ( fac [ r ] , p ) % p * modInverse ( fac [ n - r ] , p ) % p ) % p ; } function SumOfXor ( a , n ) { let mod = 10037 ; let answer = 0 ; for ( let k = 0 ; k < 32 ; k ++ ) { let x = 0 , y = 0 ; for ( let i = 0 ; i < n ; i ++ ) { if ( ( a [ i ] & ( 1 << k ) ) != 0 ) x ++ ; else y ++ ; } answer += ( ( 1 << k ) % mod * ( nCrModPFermat ( x , 3 , mod ) + x * nCrModPFermat ( y , 2 , mod ) ) % mod ) % mod ; } return answer ; } let n = 5 ; let A = [ 3 , 5 , 2 , 18 , 7 ] ; document . write ( SumOfXor ( A , n ) ) ;