Pair Duels Along a Chain

Easy

You get a singly linked list head of even length. Counting positions from zero, every even position holds an even number and every odd position holds an odd number. Split the list into neighbouring pairs: nodes 0 and 1, nodes 2 and 3, and so on. Within each pair the two numbers duel: - if the number at the odd position is larger, team "Odd" scores a point; - if the number at the even position is larger, team "Even" scores a point. Return the name of the team with more points, or "Tie" if the scores are equal.

Example 1

head=list_node([8, 3, 2, 9, 6, 1]) → 'Even'

The pairs are (8, 3), won by Even, (2, 9), won by Odd, and (6, 1), won by Even. Even wins 2 to 1.

Example 2

head=list_node([10, 11, 12, 5]) → 'Tie'

(10, 11) gives Odd a point and (12, 5) gives Even a point. It is 1 to 1, so the answer is Tie.

Example 3

head=list_node([2, 7]) → 'Odd'

Constraints

The list has between 2 and 100 nodes, and the count is even
1 ≤ node value ≤ 100
Values at even positions are even, values at odd positions are odd

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Input

head=list_node([8, 3, 2, 9, 6, 1])

Expected

'Even'

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