This is the easy version of the problem. In this version, $n \le 3000$, $x \ge y$ holds. You can make hacks only if both versions of the problem are solved.

You are given two binary strings $a$ and $b$, both of length $n$. You can do the following operation any number of times (possibly zero).

• Select two indices $l$ and $r$ ($l \lt r$).
• Change $a_l$ to $(1 - a_l)$, and $a_r$ to $(1 - a_r)$.
• If $l + 1 = r$, the cost of the operation is $x$. Otherwise, the cost is $y$.

You have to find the minimum cost needed to make $a$ equal to $b$ or say there is no way to do so.

Input

The first line contains one integer $t$ ($1 \le t \le 600$) — the number of test cases.

Each test case consists of three lines. The first line of each test case contains three integers $n$, $x$, and $y$ ($5 \le n \le 3000$, $1 \le y \le x \le 10^9$) — the length of the strings, and the costs per operation.

The second line of each test case contains the string $a$ of length $n$. The string only consists of digits $0$ and $1$.

The third line of each test case contains the string $b$ of length $n$. The string only consists of digits $0$ and $1$.

It is guaranteed that the sum of $n$ over all test cases doesn't exceed $3000$.

Output

For each test case, if there is no way to make $a$ equal to $b$, print $-1$. Otherwise, print the minimum cost needed to make $a$ equal to $b$.

Note

In the first test case, selecting indices $2$ and $3$ costs $8$, which is the minimum possible cost.

In the second test case, we cannot make $a$ equal to $b$ using any number of operations.

In the third test case, we can perform the following operations:

• Select indices $3$ and $6$. It costs $3$, and $a$ is 0101011 now.
• Select indices $4$ and $6$. It costs $3$, and $a$ is 0100001 now.

The total cost is $6$.

In the fourth test case, we don't have to perform any operations.