After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer qq independent queries.
Let's see the following example: [1,3,4][1,3,4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob — then Alice has 44 candies, and Bob has 44 candies.
Another example is [1,10,100][1,10,100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 5454 candies, and Alice takes 4646 candies. Now Bob has 5555 candies, and Alice has 5656 candies, so she has to discard one candy — and after that, she has 5555 candies too.
The first line of the input contains one integer qq (1≤q≤10001≤q≤1000) — the number of queries. Then qq queries follow.
The only line of the query contains three integers a,ba,b and cc (1≤a,b,c≤10161≤a,b,c≤1016) — the number of candies in the first, second and third piles correspondingly.
Print qq lines. The ii-th line should contain the answer for the ii-th query — the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
4 1 3 4 1 10 100 10000000000000000 10000000000000000 10000000000000000 23 34 45
4 55 15000000000000000 51
#include <bits/stdc++.h> using namespace std; int main() { int t;long long a,b,c; cin>>t; while(t--) { cin>>a>>b>>c; cout<<(a+b+c)/2<<endl; } return 0; }