In this **HackerEarth Pro and Con List problem solution,** There is a long list of n girls in front of Barney, and he is to calculate the optimal "happiness" he can find by selecting exactly 2 girls. (Why 2? No one knows!)

Ted, as a fan of pros and cons, suggests to make a list, a method for estimating the maximum happiness that Barney can achieve.

Each girl is characterized by two parameters:

- favour: if this girl is chosen, his happiness increases by this amount.

- anger: if this girl is not chosen, his happiness decreases by this amount.

Find the maximum "happiness" that Barney can obtain. Note that the answer is allowed to be negative.

## HackerEarth Pro and Con List problem solution.

#include <bits/stdc++.h>
using namespace std;
int main ()
{
int tc;
vector < int > favor;
vector < int > anger;
vector < long long int > ans;
scanf("%d",&tc);
assert(1<=tc);
assert(tc<=10);
while (tc--)
{
int n;
favor.clear();
anger.clear();
ans.clear();
scanf("%d",&n);
assert(2<=n);
assert(n<=100000);
favor.resize(n);
anger.resize(n);
ans.resize(n);
long long int hatao = 0;
for (int i=0; i<n; i++)
{
scanf("%d %d",&favor[i], &anger[i]);
ans[i] = favor[i] + anger[i];
hatao = hatao + anger[i];
assert(0<=favor[i]);
assert(favor[i]<=1000000000);
assert(0<=anger[i]);
assert(anger[i]<=1000000000);
}
sort(ans.rbegin(),ans.rend());
cout << ans[0] + ans[1] - hatao << endl;
}
return 0;
}