In a linked list of size `n`

, where `n`

is **even**, the `i`

node (^{th}**0-indexed**) of the linked list is known as the **twin** of the `(n-1-i)`

node, if ^{th}`0 <= i <= (n / 2) - 1`

.

- For example, if
`n = 4`

, then node`0`

is the twin of node`3`

, and node`1`

is the twin of node`2`

. These are the only nodes with twins for`n = 4`

.

The **twin sum **is defined as the sum of a node and its twin.

Given the `head`

of a linked list with even length, return *the maximum twin sum of the linked list*.

**Example 1:**

Input:head = [5,4,2,1]Output:6Explanation:Nodes 0 and 1 are the twins of nodes 3 and 2, respectively. All have twin sum = 6. There are no other nodes with twins in the linked list. Thus, the maximum twin sum of the linked list is 6.

**Example 2:**

Input:head = [4,2,2,3]Output:7Explanation:The nodes with twins present in this linked list are: - Node 0 is the twin of node 3 having a twin sum of 4 + 3 = 7. - Node 1 is the twin of node 2 having a twin sum of 2 + 2 = 4. Thus, the maximum twin sum of the linked list is max(7, 4) = 7.

**Example 3:**

Input:head = [1,100000]Output:100001Explanation:There is only one node with a twin in the linked list having twin sum of 1 + 100000 = 100001.

**Constraints:**

- The number of nodes in the list is an
**even**integer in the range`[2, 10`

.^{5}] `1 <= Node.val <= 10`

^{5}

` ````
```class Solution {
public:
int pairSum(ListNode* head) {
if(!head)return 0;
// either reverse the linkedList and calculate or
// save it in a vector and calculate
vector v;
while(head)
{
v.push_back(head->val);
head=head->next;
}
if(v.size()==1)return v[0];
int i=0,j=v.size()-1;
int maxSum=INT_MIN;
while(i

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