**Problem Name** : **Bitwise and Sum**

**Description** : You are given an Array A of length N. Consider that the array is one indexed.

You need to find sum of all (A[i] \land A[j] \land (A[i]+A[j])) for all pairs such that 1 \leq i < j \leq N. where \land represents bitwise and operation.

**Problem Contrains**:

1 \leq N \leq 2 \times 10^5

1\leq A[i] \leq 10^9

**Input Format**: The only argument contains the the integer array A.

**Output Format**: Return the sum of (A[i] \land A[j] \land (A[i]+A[j])) for all pairs. Since the answer can be large, return it modulo 10^9 + 7.

**Example Input**:

Input 1:

`A : [8, 9]`

Input 2:

`A : [1, 3, 3]`

**Example Output**:

Output 1:

`0`

Output 2:

`2`

**Example Explaination**:

Explaination 1:

`There is only one pair - (8 & 9 & (8 + 9)) = 0`

Explaination 2:

`We have 3 pairs -`

`(1, 3) = 1 & 3 & (1+3) = 0`

`(1, 3) = 1 & 3 & (1+3) = 0`

`(3, 3) = 3 & 3 & (3+3) = 2 `

Contest Link : https://www.interviewbit.com/contest/code-drift-bit-manipulation/