Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<ll, int> pli;
typedef pair<double,double> pdd;
#define SQ(i) ((i)*(i))
#define MEM(a, b) memset(a, (b), sizeof(a))
#define SZ(i) int(i.size())
#define FOR(i, j, k, in) for (int i=j ; i<(k) ; i+=in)
#define RFOR(i, j, k, in) for (int i=j ; i>=(k) ; i-=in)
#define REP(i, j) FOR(i, 0, j, 1)
#define REP1(i,j) FOR(i, 1, j+1, 1)
#define RREP(i, j) RFOR(i, j, 0, 1)
#define ALL(_a) _a.begin(),_a.end()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define X first
#define Y second
#ifdef tmd
#define TIME(i) Timer i(#i)
#define debug(...) do{\
    fprintf(stderr,"%s - %d (%s) = ",__PRETTY_FUNCTION__,__LINE__,#__VA_ARGS__);\
    _do(__VA_ARGS__);\
}while(0)
template<typename T>void _do(T &&_x){cerr<<_x<<endl;}
template<typename T,typename ...S> void _do(T &&_x,S &&..._t){cerr<<_x<<" ,";_do(_t...);}
template<typename _a,typename _b> ostream& operator << (ostream &_s,const pair<_a,_b> &_p){return _s<<"("<<_p.X<<","<<_p.Y<<")";}
template<typename It> ostream& _OUTC(ostream &_s,It _ita,It _itb)
{
    _s<<"{";
    for(It _it=_ita;_it!=_itb;_it++)
    {
        _s<<(_it==_ita?"":",")<<*_it;
    }
    _s<<"}";
    return _s;
}
template<typename _a> ostream &operator << (ostream &_s,vector<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,set<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a> ostream &operator << (ostream &_s,deque<_a> &_c){return _OUTC(_s,ALL(_c));}
template<typename _a,typename _b> ostream &operator << (ostream &_s,map<_a,_b> &_c){return _OUTC(_s,ALL(_c));}
template<typename _t> void pary(_t _a,_t _b){_OUTC(cerr,_a,_b);cerr<<endl;}
#define IOS()
class Timer {
private:
    string scope_name;
    chrono::high_resolution_clock::time_point start_time;
public:
    Timer (string name) : scope_name(name) {
        start_time = chrono::high_resolution_clock::now();
    }
    ~Timer () {
        auto stop_time = chrono::high_resolution_clock::now();
        auto length = chrono::duration_cast<chrono::microseconds>(stop_time - start_time).count();
        double mlength = double(length) * 0.001;
        debug(scope_name, mlength);
    }
};
#else
#define TIME(i)
#define debug(...)
#define pary(...)
#define endl '\n'
#define IOS() ios_base::sync_with_stdio(0);cin.tie(0)
#endif

const ll MOD = 1000000007;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int iNF = 0x3f3f3f3f;
 const ll MAXN = 200005;

int n, a[MAXN], b[MAXN];
ll ans = 0;
/********** Good Luck :) **********/
int main () {
    TIME(main);
    IOS();

    cin >> n;


    int ecnt = 0;

    vector<int> gt;
    REP (i, n) {
        cin >> a[i] >> b[i];
        if (a[i] < b[i]) {
            ans += b[i];
        }
        if (a[i] == b[i]) {
            gt.emplace_back(a[i]);
        }
    }

    sort(gt.rbegin(), gt.rend());
    REP (i, SZ(gt)) {
        if (i%2 == 0) {
            ans += gt[i];
        }
    }

    cout << (SZ(gt) == n? 0 : ans) << endl;

    return 0;
}