A. Winner

The winner of the card game popular in Berland “Berlogging” is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line “name score”, where name is a player’s name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to m) at the end of the game, than wins the one of them who scored at least m points first. Initially each player has 0 points. It’s guaranteed that at the end of the game at least one player has a positive number of points.

Input

The first line contains an integer number n (1  ≤  n  ≤  1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in “name score” format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.

Output

Print the name of the winner.

Examples

input

3
mike 3
andrew 5
mike 2

output

andrew

input

3
andrew 3
andrew 2
mike 5

output

andrew

---

思路：

该题求解玩家的最高分获得者，需要说明的一点是：如果有最高分相同的玩家，即最终得分都是最高分，那么其中最先达到或超过最高分（score >= max， 原题说明who scored at least m points first）的玩家获胜。

该题由于可能要求得最先达到最高分的玩家，因此使用两次计算的方法，第一次先求得所有玩家的最终得分，并从中选取最高分；第二次为了找到最先达到最高分的玩家，需要再计算一遍得分过程，当score>=max时，便是最终答案。

坑点：由于会有扣分情况，即出现负数，所以不能直接一次循环比较当前max值得到最高分，因为在扣分后，max可能不再满足是当前选手中的成绩（例如，max是100，刚好该选手现在被扣分，成了90，但是max的值并未被更新为当前所有成绩的最大值）

---

code：

#include <iostream>
#include <string>
#include <map>
using namespace std;int main() {map< string, int > _map;  //记录所有玩家最终成绩map< string, int > _map2;  //二次计算得最先达到最高分的玩家string name[1000];int score[1000], n, max = -99999999;cin >> n;for( int i = 0; i < n; i++ ) {cin >> name[i] >> score[i];_map[ name[i] ] += score[i];}/* 获得最高分 */map< string, int >::iterator it;for( it = _map.begin(); it != _map.end(); it++ ) {if ( max < it->second ) {max = it->second;}}for( int i = 0; i < n; i++ ) {if ( _map[ name[i] ] == max ) {_map2[ name[i] ] += score[i];/* 最先达到最高分 */if ( _map2[ name[i] ] >= max ) {  cout << name[i];break;}}}return 0;
}