不恋尘世浮华，不写红尘纷扰，不叹世道苍凉，不惹情思哀怨，闲看花开，静待花落，冷暖自知，干净如始。

Vasya has a pile, that consists of some number of stones. nn times he either took one stone from the pile or added one stone to the pile. The pile was non-empty before each operation of taking one stone from the pile.

You are given nn operations which Vasya has made. Find the minimal possible number of stones that can be in the pile after making these operations.

Input

The first line contains one positive integer nn — the number of operations, that have been made by Vasya (1≤n≤1001≤n≤100).

The next line contains the string ss, consisting of nn symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on ii-th operation, sisi is equal to "-" (without quotes), if added, sisi is equal to "+" (without quotes).

Output

Print one integer — the minimal possible number of stones that can be in the pile after these nnoperations.

Examples

Input

3 ---

Output

0

Input

4 ++++

Output

4

Input

2 -+

Output

1

Input

5 ++-++

Output

3

Note

In the first test, if Vasya had 33 stones in the pile at the beginning, after making operations the number of stones will be equal to 00. It is impossible to have less number of piles, so the answer is 00. Please notice, that the number of stones at the beginning can't be less, than 33, because in this case, Vasya won't be able to take a stone on some operation (the pile will be empty).

In the second test, if Vasya had 00 stones in the pile at the beginning, after making operations the number of stones will be equal to 44. It is impossible to have less number of piles because after making 44 operations the number of stones in the pile increases on 44 stones. So, the answer is 44.

In the third test, if Vasya had 11 stone in the pile at the beginning, after making operations the number of stones will be equal to 11. It can be proved, that it is impossible to have less number of stones after making the operations.

In the fourth test, if Vasya had 00 stones in the pile at the beginning, after making operations the number of stones will be equal to 33.

#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<string> #include<map> #include<algorithm> #include<iomanip> #include<queue> #include<set> #include<stack> using namespace std; int main() { int n,i,sum=0; char s[105]; cin>>n; for(i=0;i<n;i++) { cin>>s[i]; if(s[i]=='+') sum++; else { sum--; if(sum<0) sum=0; } } cout<<sum<<endl; return 0; }