1159_A. A pile of stones
Vasya has a pile, that consists of some number of stones. n 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 n 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 n — the number of operations, that have been made by Vasya (1 ≤ n ≤ 100).

The next line contains the string s, consisting of n symbols, equal to "-" (without quotes) or "+" (without quotes). If Vasya took the stone on i-th operation, s_i is equal to "-" (without quotes), if added, s_i is equal to "+" (without quotes).

Output

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

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 3 stones in the pile at the beginning, after making operations the number of stones will be equal to 0. It is impossible to have less number of piles, so the answer is 0. Please notice, that the number of stones at the beginning can't be less, than 3, 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 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 4. It is impossible to have less number of piles because after making 4 operations the number of stones in the pile increases on 4 stones. So, the answer is 4.

In the third test, if Vasya had 1 stone in the pile at the beginning, after making operations the number of stones will be equal to 1. 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 0 stones in the pile at the beginning, after making operations the number of stones will be equal to 3.
```python
'''
t= input()
lng= len(t)
li1=[]; li2=[]
for i in range(lng):
    if t[i]!='a':
        li1.append(t[i])
    elif t[i]=='a':
        li2.append(i)
        
aa= ''.join(li1)
if len(aa)==0:
    print(t); exit(0)
if len(aa)%2==1:
    print(':('); exit(0)
if len(aa)%2==0:
    #print(123)
    l= int(len(aa)/2); lp= l#; print(l)
    ll= aa[l:]; lll=aa[0:l] #; print(ll)
    if ll!=lll:
        print(':('); exit(0)
    if ll not in t:
        print(':('); exit(0)
    tp= t[::-1]; tc= ll[::-1]#; print(tp,tc)
    if tp.find(tc)!=0:
        print(':('); exit(0)
    if tp.find(tc)==0:
        ul= len(tc)
        lu= tp[ul:][::-1]
        print(lu); exit(0)
        
   '''

n= int(input()); s= input(); ln= len(s); sv=0
for i in s:
    if i=='+':
        sv+=1
    if i=='-' and sv>0:
        sv-=1
print(sv)
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
        
```