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

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思路：

該題求解玩家的最高分獲得者，需要說明的一點是：如果有最高分相同的玩家，即最終得分都是最高分，那么其中最先達到或超過最高分（score >= max， 原題說明who scored at least m points first）的玩家獲勝。

該題由于可能要求得最先達到最高分的玩家，因此使用兩次計算的方法，第一次先求得所有玩家的最終得分，并從中選取最高分；第二次為了找到最先達到最高分的玩家，需要再計算一遍得分過程，當score>=max時，便是最終答案。

坑點：由于會有扣分情況，即出現負數，所以不能直接一次循環比較當前max值得到最高分，因為在扣分后，max可能不再滿足是當前選手中的成績（例如，max是100，剛好該選手現在被扣分，成了90，但是max的值并未被更新為當前所有成績的最大值）

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code：

#include <iostream> #include <string> #include <map> using namespace std;int main() {map< string, int > _map; //記錄所有玩家最終成績map< string, int > _map2; //二次計算得最先達到最高分的玩家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];}/* 獲得最高分 */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];/* 最先達到最高分 */if ( _map2[ name[i] ] >= max ) { cout << name[i];break;}}}return 0; }

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