We have an integer sequence $A$ of length $N$ and an integer sequence $B$ of length $M$.  
Takahashi will make a new sequence $A'$ by removing some elements (possibly zero or all) from $A$ and concatenating the remaining elements.  
Similarly, he will make another new sequence $B'$ by removing some elements (possibly zero or all) from $B$ and concatenating the remaining elements.  
Here, he will remove elements so that $|A'| = |B'|$. ($|s|$ denotes the length of $s$ for a sequence $s$.)  
Let $x$ be the total number of elements removed from $A$ and $B$, and $y$ be the number of integers $i$ such that $1 \le i \le |A'|$ and ${A'}_i \neq {B'}_i$. Print the minimium possible value of $x + y$.

Input is given from Standard Input in the following format:

```
$N$ $M$
$A_1 \hspace{7pt} A_2 \hspace{7pt} A_3 \hspace{5pt} \dots \hspace{5pt} A_N$
$B_1 \hspace{7pt} B_2 \hspace{7pt} B_3 \hspace{5pt} \dots \hspace{5pt} B_M$
```

Print the minimum possible value of $x + y$.

-   $1 \le N, M \le 1000$
-   $1 \le A_i, B_i \le 10^9$
-   All values in input are integers.

2

If we make A′ by removing $A_4$ from A, and B′ by removing nothing from B, x will be 1.
Here, there is just one integer i such that 1≤i≤∣A′∣ and $A′_i≠B′_i:\ i=2$, so y will be 1, and x+y will be 2, which is the minimum possible value.

3

If we remove nothing from A and remove $B_4,B_6$ from B, we have x=2,y=1, and x+y=3, which is the minimum possible value.

5

It is allowed to remove nothing from both A and B.