Codeforces 282 a Bit++ problem solution

The classic programming language of Bitland is Bit++. This language is so peculiar and complicated.

The language is that peculiar as it has exactly one variable, called x. Also, there are two operations:

1. Operation ++ increases the value of variable x by 1.
2. Operation — decreases the value of variable x by 1.

A statement in language Bit++ is a sequence, consisting of exactly one operation and one variable x. The statement is written without spaces, that is, it can only contain characters ” +”, ”-”, “X”. Executing a statement means applying the operation it contains.

A programme in Bit++ is a sequence of statements, each of them needs to be executed. Executing a programme means executing all the statements it contains.

You’re given a programme in language Bit++. The initial value of x is 0. Execute the programme and find its final value (the value of the variable when this programme is executed).

Input

The first line contains a single integer n (1 ≤ n ≤ 150) — the number of statements in the programme.

Next n lines contain a statement each. Each statement contains exactly one operation (++ or —) and exactly one variable x (denoted as letter «X»). Thus, there are no empty statements. The operation and the variable can be written in any order.

Output

Print a single integer — the final value of x.

Examples

input

1
++X

output

1

input
2
X++
--X

output

0

Codeforces 282a Bit++ problem solution in java

// 282a bit++
import java.util.Scanner;

public class Bit {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int x = 0;
        int num_of_cases = sc.nextInt();
        while (num_of_cases-- > 0) {
            String s = sc.next();
            if (s.equals("++X") || s.equals("X++"))
                x++;
            else
                x--;
        }
        System.out.println(x);
        sc.close();
    }
}