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.

/*
9 2
        1 1 1 1 1 0 0 0 0
        1 2
        1 5
        5 6
        5 7
        2 3
        2 4
        3 8
        3 9
*/



import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.net.CookieHandler;
import java.util.*;

public class Question{



    public static  void main(String[] args) throws IOException {

        Reader.init(System.in);
        int n = Reader.nextInt();
        String x = Reader.next();

        char[] arr = x.toCharArray();
        int ans =0 ;
        for (int i =  0 ;i < n ; i++){
            if (arr[i]=='-'){
                ans--;
            }
            else{
                ans++;
            }

            ans = Math.max(0,ans);
        }

        System.out.println(ans);


























    }

    static long fun(long n){
        long max = -1;
        for (int i = 0 ; i < n ; i++){
            String x = Integer.toString(i);
            long product = 1;
            for (int j = 0 ; j < x.length() ; j++){
                product = product * Long.parseLong(Character.toString(x.charAt(j)));
            }
            max = Math.max(product,max);
        }
        return max;
    }




}



class Reader {
    static BufferedReader reader;
    static StringTokenizer tokenizer;

    /** call this method to initialize reader for InputStream */
    static void init(InputStream input) {
        reader = new BufferedReader(
                new InputStreamReader(input) );
        tokenizer = new StringTokenizer("");
    }

    /** get next word */
    static String next() throws IOException {
        while ( ! tokenizer.hasMoreTokens() ) {
            //TODO add check for eof if necessary
            tokenizer = new StringTokenizer(
                    reader.readLine() );
        }
        return tokenizer.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt( next() );
    }
    static long nextLong() throws IOException {
        return Long.parseLong( next() );
    }

    static double nextDouble() throws IOException {
        return Double.parseDouble( next() );
    }
}

class MergeSort {
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];


        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);

            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    /* A utility function to print array of size n */
    static void printArray(int arr[]) {
        int n = arr.length;
        for (int i = 0; i < n; ++i)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    static long phi(long n,long R)
    {

        long result = R;


        for (long p = 2; p * p <= n; ++p)
        {


            if (n % p == 0)
            {


                while (n % p == 0)
                    n =n/p;
                result =result - ( result / p);
            }
        }


        //System.out.println("n" + n);
        if (n > 1)
            result -= result / n;
        return result;
    }
}
