题干：

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

题目大意：

给出一些名字和分数，正的表示加分，负的表示减分，要求你求出最大分数而且最先得到这个最高分的那个人的名字

解题报告：

  这题坑还是很多的。。你不能在线维护一个最大值，因为这个最大值还可能减下来，，你需要看的是最后比赛完了以后的最大值，记录那个最大值。并且最后判断输出的时候也不能直接找到第一个大于maxx的就停止搜索了，因为这个人也可能后来有负分导致最终分数小于maxx。

AC代码：

#include<bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
map<string, int> mp;
int main()
{
	int n,sc[1000 + 5];//sc代表分数 
	char s[1000 + 5][50];
	int maxx = -INF;
	cin>>n;
	for(int i = 1; i<=n; i++) {
		scanf("%s %d",s[i],&sc[i]);
		mp[s[i]]+=sc[i];
//		maxx =max(maxx,mp[s[i]]); 
	}
	for(int i = 1; i<=n; i++) {
		maxx = max(maxx,mp[s[i]]);
	}
	map<string,int > mpp;
	for(int i = 1; i<=n; i++) {
		mpp[s[i]] +=sc[i];
		if(mpp[s[i]] >= maxx &&mp[s[i]] == maxx ) {
			printf("%s\n",s[i]);
			break; 
		} 
	}
	
	return 0 ;
 }