One very experienced problem writer decided to prepare a problem for April Fools Day contest. The task was very simple - given an arithmetic expression, return the result of evaluating this expression. However, looks like there is a bug in the reference solution...
## Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
## Output
Reproduce the output of the reference solution, including the bug.
[samples]
一位非常有经验的题目出题人决定为愚人节比赛准备一道题。任务非常简单——给定一个算术表达式，返回该表达式的计算结果。然而，参考程序似乎存在一个 bug...
输入数据仅有一行，包含一个算术表达式。表达式将包含 2 到 10 个操作数，由加号和/或减号分隔。每个操作数均为 0 到 255（含）之间的整数。
请复现参考程序的输出，包括其 bug。
## Input
The only line of input data contains the arithmetic expression. The expression will contain between 2 and 10 operands, separated with arithmetic signs plus and/or minus. Each operand will be an integer between 0 and 255, inclusive.
## Output
Reproduce the output of the reference solution, including the bug.
[samples]
**Definitions**
Let $ E $ be an arithmetic expression of the form:
$$
a_1 \circ_1 a_2 \circ_2 a_3 \circ_3 \cdots \circ_{n-1} a_n
$$
where:
- $ n \in \mathbb{Z} $, $ 2 \leq n \leq 10 $, is the number of operands,
- $ a_i \in \mathbb{Z} $, $ 0 \leq a_i \leq 255 $, for all $ i \in \{1, \dots, n\} $,
- $ \circ_j \in \{+, -\} $ for all $ j \in \{1, \dots, n-1\} $.
**Constraints**
1. The expression is given as a single string with no parentheses, no whitespace, and operands and operators concatenated directly.
2. All operands are non-negative integers in $[0, 255]$.
3. Operators are only `+` and `-`.
**Objective**
Reproduce the output of a reference solution with the following bug:
> The expression is evaluated **left-to-right**, ignoring standard operator precedence, **but** the first operator is **always treated as addition**, regardless of its actual symbol.
That is, define the evaluation function $ f(E) $ as:
$$
f(E) = a_1 \oplus a_2 \oplus a_3 \oplus \cdots \oplus a_n
$$
where:
- $ \oplus $ denotes left-to-right binary operation,
- The operation between $ a_1 $ and $ a_2 $ is **always addition**, i.e., $ a_1 + a_2 $,
- For $ i \geq 3 $, the operation between the accumulated result and $ a_i $ is determined by $ \circ_{i-1} $:
$$
\text{result} =
\begin{cases}
\text{result} + a_i & \text{if } \circ_{i-1} = + \\
\text{result} - a_i & \text{if } \circ_{i-1} = -
\end{cases}
$$
**Note**: The bug is that the **first operator $ \circ_1 $ is ignored** and replaced with `+`, even if it is `-`.