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.
import sys
# sys.setrecursionlimit(300000)
# Get out of main functoin
def main():
    pass
# decimal to binary
def binary(n):
    return (bin(n).replace("0b", ""))
# binary to decimal
def decimal(s):
    return (int(s, 2))
# power of a number base 2
def pow2(n):
    p = 0
    while n > 1:
        n //= 2
        p += 1
    return (p)
# if  number is prime in √n time
def isPrime(n):
    if (n == 1):
        return (False)
    else:
        root = int(n ** 0.5)
        root += 1
        for i in range(2, root):
            if (n % i == 0):
                return (False)
        return (True)
# list to string ,no spaces
def lts(l):
    s = ''.join(map(str, l))
    return s
# String to list
def stl(s):
    # for each character in string to list with no spaces -->
    l = list(s)
    # for space in string  -->
    # l=list(s.split(" "))
    return l
# Returns list of numbers with a particular sum
def sq(a, target, arr=[]):
    s = sum(arr)
    if (s == target):
        return arr
    if (s >= target):
        return
    for i in range(len(a)):
        n = a[i]
        remaining = a[i + 1:]
        ans = sq(remaining, target, arr + [n])
        if (ans):
            return ans
# Sieve for prime numbers in a range
def SieveOfEratosthenes(n):
    cnt = 0
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    for p in range(2, n + 1):
        if prime[p]:
            cnt += 1
            # print(p)
    return (cnt)
# for positive integerse only
def nCr(n, r):
    f = math.factorial
    return f(n) // f(r) // f(n - r)
# 1000000007
mod = int(1e9) + 7
def ssinp(): return sys.stdin.readline().strip()
# s=input()
def iinp(): return int(input())
# n=int(input())
def nninp(): return map(int, sys.stdin.readline().strip().split())
# n, m, a=[int(x) for x in input().split()]
def llinp(): return list(map(int, sys.stdin.readline().strip().split()))
# a=list(map(int,input().split()))
def p(xyz): print(xyz)
def p2(a, b): print(a, b)
import math
#list.sort(key=lambda x:x[1]) for sorting a list according to second element in sublist
#d1.setdefault(key, []).append(value)
#ASCII of A-Z= 65-90
#ASCII of a-z= 97-122
#import random
#from collections import OrderedDict
#from fractions import Fraction
#============Speed: STRING < LIST < SET,DICTIONARY==========================
#for __ in range(iinp()):
n=iinp()
s=ssinp()
sum=ans=0
for c in s:
    if(c=="+"):
        sum+=1
    else:
        sum-=1
    ans=min(ans,sum)
p(sum-ans)
""" Stuff you should look for
    int overflow, array bounds
    special cases (n=1?)
    do something instead of nothing and stay organized
    WRITE STUFF DOWN
	DON'T GET STUCK ON ONE APPROACH"""
