CF427A.Police Recruits

The police department of your city has just started its journey. Initially, they don’t have any manpower. So, they started hiring new recruits in groups.

Meanwhile, crimes keeps occurring within the city. One member of the police force can investigate only one crime during his/her lifetime.

If there is no police officer free (isn't busy with crime) during the occurrence of a crime, it will go untreated.

Given the chronological order of crime occurrences and recruit hirings, find the number of crimes which will go untreated.

你所在城市的警察部门刚刚成立。最初，他们没有任何警力，因此开始以小组形式招聘新警员。

与此同时，城市中不断发生犯罪事件。每名警员在其职业生涯中只能调查一起犯罪案件。

如果在某起犯罪发生时，没有空闲的警员（即所有警员都正在处理其他犯罪案件），那么该犯罪将得不到处理。

给定犯罪发生与警员招聘的时间顺序，求将得不到处理的犯罪数量。

The first line of input will contain an integer n (1 ≤ n ≤ 105), the number of events. The next line will contain n space-separated integers.

If the integer is -1 then it means a crime has occurred. Otherwise, the integer will be positive, the number of officers recruited together at that time. No more than 10 officers will be recruited at a time.

输入的第一行包含一个整数 $n$（$1 \leq n \leq 10^5$），表示事件的数量。
第二行包含 $n$ 个用空格分隔的整数。

如果该整数为 $-1$，则表示发生了一起犯罪事件；否则，该整数为正数，表示在该时刻一同招募的警员数量。每次招募的警员数量不超过 $10$ 名。

Print a single integer, the number of crimes which will go untreated.

输出一个整数，表示将未被处理的犯罪数量。

Lets consider the second example:

1. Firstly one person is hired.
2. Then crime appears, the last hired person will investigate this crime.
3. One more person is hired.
4. One more crime appears, the last hired person will investigate this crime.
5. Crime appears. There is no free policeman at the time, so this crime will go untreated.
6. One more person is hired.
7. One more person is hired.
8. One more person is hired.

The answer is one, as one crime (on step 5) will go untreated.

我们考虑第二个例子：

1. 首先聘用一名人员。
2. 然后发生一起犯罪事件，最后聘用的人员将调查该犯罪事件。
3. 再聘用一名人员。
4. 又发生一起犯罪事件，最后聘用的人员将调查该犯罪事件。
5. 再发生一起犯罪事件。此时没有空闲的警员，因此该犯罪事件将得不到处理。
6. 再聘用一名人员。
7. 再聘用一名人员。
8. 再聘用一名人员。

答案为一，因为有一起犯罪事件（第 5 步）将得不到处理。