CF1433F.Zero Remainder Sum

You are given a matrix $a$ of size $n \times m$ consisting of integers.

You can choose no more than $\left\lfloor\frac{m}{2}\right\rfloor$ elements in each row. Your task is to choose these elements in such a way that their sum is divisible by $k$ and this sum is the maximum.

In other words, you can choose no more than a half (rounded down) of elements in each row, you have to find the maximum sum of these elements divisible by $k$.

Note that you can choose zero elements (and the sum of such set is $0$).

给你一个大小为 $n \times m$ 的整数矩阵 $a$。

你可以在每一行中至多选择 $\left\lfloor\frac{m}{2}\right\rfloor$ 个元素。你的任务是选择这些元素，使得它们的和能被 $k$ 整除，并且该和尽可能大。

换句话说，你可以在每一行中至多选择一半（向下取整）的元素，你需要找出满足能被 $k$ 整除的最大可能元素和。

注意：你可以选择零个元素（此时和为 $0$）。

The first line of the input contains three integers $n$, $m$ and $k$ ($1 \le n, m, k \le 70$) — the number of rows in the matrix, the number of columns in the matrix and the value of $k$. The next $n$ lines contain $m$ elements each, where the $j$-th element of the $i$-th row is $a_{i, j}$ ($1 \le a_{i, j} \le 70$).

输入的第一行包含三个整数 $n$、$m$ 和 $k$（$1 \le n, m, k \le 70$），分别表示矩阵的行数、列数以及 $k$ 的值。接下来的 $n$ 行每行包含 $m$ 个元素，其中第 $i$ 行的第 $j$ 个元素为 $a_{i, j}$（$1 \le a_{i, j} \le 70$）。

Print one integer — the maximum sum divisible by $k$ you can obtain.

输出一个整数——你能得到的能被 $k$ 整除的最大和。

In the first example, the optimal answer is $2$ and $4$ in the first row, $5$ and $2$ in the second row and $7$ and $4$ in the third row. The total sum is $2 + 4 + 5 + 2 + 7 + 4 = 24$.

在第一个例子中，最优解是第一行取 $2$ 和 $4$，第二行取 $5$ 和 $2$，第三行取 $7$ 和 $4$。总和为 $2 + 4 + 5 + 2 + 7 + 4 = 24$。