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.
// Change Of Plans BABY.... Change Of Plans //
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class aPileOfStones {
    static void MainSolution() {
        n=ni();
        char ch[]=ns().toCharArray();
        int c=0;
        for(int i=0;i<n;i++){
            if(ch[i]=='-' && c>0)c--;
            else if(ch[i]=='+') c++;
        }
        pl(c);
    }
    /*----------------------------------------------------------------------------------------------------------------*/
    //THE DON'T CARE ZONE BEGINS HERE...\\
    static int mod9 = 1000000007;
    static int n, m, l, k, t;
    static AwesomeInput input = new AwesomeInput(System.in);
    static PrintWriter pw = new PrintWriter(System.out, true);

    // The Awesome Input Code is a fast IO method using Input Stream but with certain manipulations by me //
    static class AwesomeInput {
        private InputStream letsDoIT;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        private AwesomeInput(InputStream incoming) {
            this.letsDoIT = incoming;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = letsDoIT.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        private long ForLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        private String ForString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            boolean isSpaceChar(int ch);
        }
    }

    // functions to take input//
    static int ni() {
        return (int) input.ForLong();
    }

    static String ns() {
        return input.ForString();
    }

    static long nl() {
        return input.ForLong();
    }

    static double nd() throws IOException {
        return Double.parseDouble(new BufferedReader(new InputStreamReader(System.in)).readLine());
    }

    //functions to give output
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pws(Object o) {
        pw.print(o + "");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    // Fast Sort is Radix Sort
    public static int[] fastSort(int[] f) {
        int n = f.length;
        int[] to = new int[n];
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] & 0xffff)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] & 0xffff]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] >>> 16)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] >>> 16]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        return f;
    }

    public static void main(String[] args) {      //threading has been used to increase the stack size.

        new Thread(null, null, "Vengeance", 1 << 25)  //the last parameter is stack size desired.
        {
            public void run() {
                try {
                    double s = System.currentTimeMillis();
                    MainSolution();
                    //pl(("\nExecution Time : " + ((double) System.currentTimeMillis() - s) / 1000) + " s");
                    pw.flush();
                    pw.close();
                } catch (Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }
}