B

Problem description

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:

• Operation ++ increases the value of variable x by 1.
• 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
解题思路：给X初始值为0，通过n个操作，实现加1减1运算，水过！
AC代码：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,x=0;char q[5];
    cin>>n;
    while(n--){
        cin>>q;
        if(q[0]=='+'||q[1]=='+')x++;
        else x--;
    }
    cout<<x<<endl;
    return 0;
}