Problem:
Score : 200 points

Problem Statement

A group of people played a game. All players had distinct scores, which are positive integers.

Takahashi knows N facts on the players' scores. The i-th fact is as follows: the A_i-th highest score among the players is B_i.

Find the maximum possible number of players in the game.

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9(1\leq i\leq N)

0 \leq B_i \leq 10^9(1\leq i\leq N)

If i ≠ j, A_i ≠ A_j.

There exists a possible outcome of the game that are consistent with the facts.

All input values are integers.

Inputs

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Outputs

Print the maximum possible number of players in the game.

Sample Input 1

3
4 7
2 9
6 2

Sample Output 1

8

The maximum possible number of players is achieved when, for example, the players have the following scores: 12,9,8,7,5,2,1,0.

Sample Input 2

5
1 10
3 6
5 2
4 4
2 8

Sample Output 2

7

Sample Input 3

2
1 1000000000
1000000000 1

Sample Output 3

1000000001

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
/* clang-format off */
#define MOD 1000000007
#define INF 1000000000
#define REP(i, n) for (ll i = 0, i##_len = (n); i < i##_len; ++i)
#define ALL(a) (a).begin(), (a).end()
#define __DEBUG__
#ifdef __DEBUG__
#define CH_P(a) cout <<"check_point("<<#a<<")" << "\n";
#define DEBUG(x) cout<<#x<<":"<<x<<"\n"
#define DEBUGS(v) cout << #v << ":";for(auto x:v){cout<<x<<" ";}cout<<"\n"
#endif
#ifndef __DEBUG__
#define CH_P(a) 
#define DEBUG(x) 
#define DEBUGS(v) 
#endif
#define TIMER_S(start) chrono::system_clock::time_point start = chrono::system_clock::now();
#define TIMER_E(end) chrono::system_clock::time_point end = chrono::system_clock::now();
#define TIME(start, end) cout << static_cast<double>(chrono::duration_cast<chrono::microseconds>(end - start).count()) << "ms" << "\n";
/* clang-format on */

int main() {
  ll n;
  cin>>n;
  vector<ll> a(n), b(n);
  REP(i, n) { cin >> a[i] >> b[i]; }
  ll m=-1,t;
  REP(i, n) {
    if (a[i] > m){ t = i;m=a[i];}
  }
  cout <<  a[t]+b[t]<< endl;

  return 0;
}