A student called Slon is very mischievous in school. He is always bored in class and he is always making a mess. The teacher wanted to calm him down and "tame" him, so he has given him a difficult mathematical problem.

The teacher gives Slon an arithmetic expression , the integer and . Slon has to answer the following question: "What is the **minimal non-negative** value of variable in expression so that the remainder of dividing with is equal to ?". The solution will always **exist**.

Additionally, it will hold that, if we apply the **laws of distribution** on expression , variable will not multiply variable (formally, the expression is a polynomial of the first degree in variable ).

Examples of valid expressions : .

Examples of invalid expressions : .

#### Input Specification

The first line of input contains the expression .

The second line of input contains two integers , .

The arithmetic expression will only consists of characters `+`

, `-`

, `*`

, `(`

, `)`

, `x`

and digits from `0`

to `9`

.

The brackets will always be paired, the operators `+`

, `-`

and `*`

will always be applied to exactly two values (there will not be an expression `(-5)`

or `(4+-5)`

) and all multiplications will be explicit (there will not be an expression `4(5)`

or `2(x)`

).

#### Output

The first and only line of output must contain the minimal non-negative value of variable .

#### Sample Input 1

```
5+3+x
9 10
```

#### Sample Output 1

`1`

#### Explanation for Sample Output 1

The remainder of dividing with for is , and the remainder of division for is , which is the solution.

#### Sample Input 2

```
20+3+x
0 5
```

#### Sample Output 2

`2`

#### Sample Input 3

```
3*(x+(x+4)*5)
1 7
```

#### Sample Output 3

`1`

## Comments

does BEDMAS apply?

Yes.

ok

There are two new cases (case 11 and 12), courtesy of

d. Each case is worth 25% of all points.I asked

dwhy he wanted to add extra test cases and he told me: