CF1294E.Obtain a Permutation

You are given a rectangular matrix of size $n \times m$ consisting of integers from $1$ to $2 \cdot 10^5$.

In one move, you can:

• choose any element of the matrix and change its value to any integer between $1$ and $n \cdot m$, inclusive;
• take any column and shift it one cell up cyclically (see the example of such cyclic shift below).

A cyclic shift is an operation such that you choose some $j$ ($1 \le j \le m$) and set $a_{1, j} := a_{2, j}, a_{2, j} := a_{3, j}, \dots, a_{n, j} := a_{1, j}$ simultaneously.

Example of cyclic shift of the first column

You want to perform the minimum number of moves to make this matrix look like this:

In other words, the goal is to obtain the matrix, where $a_{1, 1} = 1, a_{1, 2} = 2, \dots, a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, \dots, a_{n, m} = n \cdot m$ (i.e. $a_{i, j} = (i - 1) \cdot m + j$) with the minimum number of moves performed.

你被给定一个大小为 $n \times m$ 的矩形矩阵，其中元素为 $1$ 到 $2 \cdot 10^5$ 之间的整数。

在一次操作中，你可以执行以下任一动作：

• 选择矩阵中的任意一个元素，并将其值更改为 $1$ 到 $n \cdot m$（含）之间的任意整数；
• 选择任意一列，并将其循环上移一格（如下图所示为该循环移位操作的示例）。

循环移位是指：选定某个 $j$（$1 \le j \le m$），同时执行如下赋值：$a_{1, j} := a_{2, j},\; a_{2, j} := a_{3, j},\; \dots,\; a_{n, j} := a_{1, j}$。

第一列循环上移的示例

你的目标是通过最少的操作次数，将该矩阵变为如下形式：

换言之，最终矩阵需满足：$a_{1, 1} = 1,\; a_{1, 2} = 2,\; \dots,\; a_{1, m} = m,\; a_{2, 1} = m + 1,\; a_{2, 2} = m + 2,\; \dots,\; a_{n, m} = n \cdot m$（即对任意 $i,j$，有 $a_{i, j} = (i - 1) \cdot m + j$），且操作总次数应最小。

The first line of the input contains two integers $n$ and $m$ ($1 \le n, m \le 2 \cdot 10^5, n \cdot m \le 2 \cdot 10^5$) — the size of the matrix.

The next $n$ lines contain $m$ integers each. The number at the line $i$ and position $j$ is $a_{i, j}$ ($1 \le a_{i, j} \le 2 \cdot 10^5$).

输入的第一行包含两个整数 $n$ 和 $m$（$1 \le n, m \le 2 \cdot 10^5$，且 $n \cdot m \le 2 \cdot 10^5$）——表示矩阵的大小。

接下来的 $n$ 行，每行包含 $m$ 个整数。第 $i$ 行第 $j$ 列的数字为 $a_{i, j}$（$1 \le a_{i, j} \le 2 \cdot 10^5$）。

Print one integer — the minimum number of moves required to obtain the matrix, where $a_{1, 1} = 1, a_{1, 2} = 2, \dots, a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, \dots, a_{n, m} = n \cdot m$ ($a_{i, j} = (i - 1)m + j$).

输出一个整数——得到如下矩阵所需的最少移动次数：其中 $a_{1, 1} = 1, a_{1, 2} = 2, \dots, a_{1, m} = m, a_{2, 1} = m + 1, a_{2, 2} = m + 2, \dots, a_{n, m} = n \cdot m$（即 $a_{i, j} = (i - 1)m + j$）。

In the first example, you can set $a_{1, 1} := 7, a_{1, 2} := 8$ and $a_{1, 3} := 9$ then shift the first, the second and the third columns cyclically, so the answer is $6$. It can be shown that you cannot achieve a better answer.

In the second example, the matrix is already good so the answer is $0$.

In the third example, it is enough to shift the second column cyclically twice to obtain a good matrix, so the answer is $2$.

在第一个例子中，你可以设置 $a_{1, 1} := 7, a_{1, 2} := 8$ 和 $a_{1, 3} := 9$，然后对第一、第二和第三列分别进行循环移位，因此答案为 $6$。可以证明无法得到更优的答案。

在第二个例子中，矩阵本身已是“好”矩阵，因此答案为 $0$。

在第三个例子中，只需将第二列循环移位两次即可得到一个“好”矩阵，因此答案为 $2$。