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 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

/* define */
#define Yes(X) puts ( X ? "Yes" : "No")
#define YES(X) puts ( X ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = a; i < b ; i++)
#define REP(i, n) for ( int i = 0; i < n ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = (int)1e9 + 7;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b); 
	return a;
}
int lcm(int a, int b) {
	return a / gcd(a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

/* main */
int main() {
	int n, a, b, ma = 1e9, mb;
	scanf("%d", &n);
	REP(i, n) {
		scanf("%d%d", &a, &b);
		if (ma < a) {
			ma = a;
			mb = b;
		}
	}
	printf("%d", ma + mb);
	return 0;
}