Codeforces Round 546 (Div. 2) |
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Finished |
Nastya came to her informatics lesson, and her teacher who is, by the way, a little bit famous here gave her the following task.
Two matrices $$$A$$$ and $$$B$$$ are given, each of them has size $$$n \times m$$$. Nastya can perform the following operation to matrix $$$A$$$ unlimited number of times:
Nastya's task is to check whether it is possible to transform the matrix $$$A$$$ to the matrix $$$B$$$.
As it may require a lot of operations, you are asked to answer this question for Nastya.
A square submatrix of matrix $$$M$$$ is a matrix which consist of all elements which comes from one of the rows with indeces $$$x, x+1, \dots, x+k-1$$$ of matrix $$$M$$$ and comes from one of the columns with indeces $$$y, y+1, \dots, y+k-1$$$ of matrix $$$M$$$. $$$k$$$ is the size of square submatrix. In other words, square submatrix is the set of elements of source matrix which form a solid square (i.e. without holes).
The first line contains two integers $$$n$$$ and $$$m$$$ separated by space ($$$1 \leq n, m \leq 500$$$) — the numbers of rows and columns in $$$A$$$ and $$$B$$$ respectively.
Each of the next $$$n$$$ lines contains $$$m$$$ integers, the $$$j$$$-th number in the $$$i$$$-th of these lines denotes the $$$j$$$-th element of the $$$i$$$-th row of the matrix $$$A$$$ ($$$1 \leq A_{ij} \leq 10^{9}$$$).
Each of the next $$$n$$$ lines contains $$$m$$$ integers, the $$$j$$$-th number in the $$$i$$$-th of these lines denotes the $$$j$$$-th element of the $$$i$$$-th row of the matrix $$$B$$$ ($$$1 \leq B_{ij} \leq 10^{9}$$$).
Print "YES" (without quotes) if it is possible to transform $$$A$$$ to $$$B$$$ and "NO" (without quotes) otherwise.
You can print each letter in any case (upper or lower).
2 2 1 1 6 1 1 6 1 1
YES
2 2 4 4 4 5 5 4 4 4
NO
3 3 1 2 3 4 5 6 7 8 9 1 4 7 2 5 6 3 8 9
YES
Consider the third example. The matrix $$$A$$$ initially looks as follows.
$$$$$$ \begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix} $$$$$$
Then we choose the whole matrix as transposed submatrix and it becomes
$$$$$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 8\\ 3 & 6 & 9 \end{bmatrix} $$$$$$
Then we transpose the submatrix with corners in cells $$$(2, 2)$$$ and $$$(3, 3)$$$.
$$$$$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & \textbf{5} & \textbf{8}\\ 3 & \textbf{6} & \textbf{9} \end{bmatrix} $$$$$$
So matrix becomes
$$$$$$ \begin{bmatrix} 1 & 4 & 7\\ 2 & 5 & 6\\ 3 & 8 & 9 \end{bmatrix} $$$$$$
and it is $$$B$$$.
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