output_description: Print the maximum value taken by x during the operations.
  
  * * *
input_description: The input is given from Standard Input in the following format:
  
      
      
      N
      S
attempt: import sys
  
  input = sys.stdin.readline
  sys.setrecursionlimit(10**7)
  from collections import Counter, deque
  from collections import defaultdict
  from itertools import combinations, permutations, accumulate, groupby, product
  from bisect import bisect_left, bisect_right
  from heapq import heapify, heappop, heappush
  from math import floor, ceil
  from operator import itemgetter
  
  
  def I():
      return int(input())
  
  
  def MI():
      return map(int, input().split())
  
  
  def LI():
      return list(map(int, input().split()))
  
  
  def LI2():
      return [int(input()) for i in range(n)]
  
  
  def MXI():
      return [[LI()] for i in range(n)]
  
  
  def printns(x):
      print("\n".join(x))
  
  
  def printni(x):
      print("\n".join(list(map(str, x))))
  
  
  inf = 10**17
  mod = 10**9 + 7
  # s=input().rstrip()
  
  sm = 0
  n = I()
  mx = 0
  s = input().rstrip()
  for i in range(n):
      if s[i] == "I":
          sm += 1
      else:
          sm -= 1
      if sm > mx:
          mx = sm
  print(mx)
  
problem_description: Statement
  
  You have an integer variable x. Initially, x=0.
  
  Some person gave you a string S of length N, and using the string you
  performed the following operation N times. In the i-th operation, you
  incremented the value of x by 1 if S_i=`I`, and decremented the value of x by
  1 if S_i=`D`.
  
  Find the maximum value taken by x during the operations (including before the
  first operation, and after the last operation).
samples: [{"input": "5\n    IIDID", "output": "2\n    \n\nAfter each operation, the value of x becomes 1, 2, 1, 2 and 1, respectively.\nThus, the output should be 2, the maximum value.\n\n* * *"}, {"input": "7\n    DDIDDII", "output": "0\n    \n\nThe initial value x=0 is the maximum value taken by x, thus the output should\nbe 0."}]