Famous Choreographer

Let's strip the ballet lore down to geometry. The dancers are just a rectangle with commitment issues.

Geometry

Suppose we add $u$ rows above the original grid, $d$ rows below it, $l$ columns to the left, and $r$ columns to the right.

The original grid stays as one contiguous $n$ by $m$ block inside the final rectangle. After a 180-degree rotation, that original block lands on another $n$ by $m$ block. Relative to the original block, the rotated copy is shifted by

$x=d-u, \qquad y=r-l.$

For a fixed $x$, the cheapest way to get $d-u=x$ uses exactly $|x|$ added rows. Same for columns. So if shift $(x,y)$ is possible, its best cost is

$(n+|x|)(m+|y|)-nm.$

Now the whole problem is: which shifts $(x,y)$ are possible?

Forced Overlaps

For shift $(x,y)$, an original cell $(i,j)$ rotates to

$(n-1+x-i,\; m-1+y-j).$

If that position is outside the original grid, no issue. We can invite a new ballerina there with the required letter.

If that position is inside the original grid, then we are stuck with the existing letter. So the two letters must match. If not, this shift is impossible. Extra padding cannot fix a contradiction between two already-existing cells. Brutal, but fair.

The statement also requires at least one original ballerina to land on an original position. That means the original block and its rotated copy must overlap, so

$-(n-1) \le x \le n-1, \qquad -(m-1) \le y \le m-1.$

The overlap size is then

$(n-|x|) \text{ by } (m-|y|).$

Subrectangle Equality

Let $B$ be the 180-degree rotation of $A$:

$B[i][j] = A[n-1-i][m-1-j].$

The condition above becomes

$A[i][j] = B[i-x][j-y]$

for every overlapping cell.

So for every shift, we only need to compare one subrectangle of $A$ with one subrectangle of $B$.

For rows, the subrectangle in $A$ starts at $max(0,x)$ and has height $n-|x|$. The matching subrectangle in $B$ starts at $max(0,x)-x$. Columns work the same way with $y$.

Trying Every Shift Is Fine

The number of shifts is

$(2n-1)(2m-1) < 4nm.$

Since the total $nm$ over all test cases is at most $10^6$, this is completely fine. The only thing we need is fast overlap comparison; checking cells one by one for every shift would be a very expensive way to be sad.

2D Hashing

Build a 2D rolling hash for $A$ and for $B$. For a matrix $M$, define prefix hash $P$ by

$P[i+1][j+1] = M[i][j] + R P[i][j+1] + C P[i+1][j] - RC P[i][j].$

Then the normalized hash of the rectangle starting at $(r,c)$ with height $h$ and width $w$ is

$P[r+h][c+w] - R^h P[r][c+w] - C^w P[r+h][c] + R^h C^w P[r][c].$

This hash is independent of the rectangle's absolute position, so equal subrectangles get equal hashes. The solution uses two independent unsigned 64-bit hashes. In contest practice, that is plenty; if you want deterministic-looking arithmetic, two large prime moduli are a drop-in replacement.

Algorithm

1. Build hash tables for $A$ and for its 180-degree rotation $B$.
2. Enumerate all $x$ from $1-n$ to $n-1$.
3. Enumerate all $y$ from $1-m$ to $m-1$.
4. Compare the overlap subrectangle of $A$ with the corresponding subrectangle of $B$.
5. If they match, update the answer with $(n+|x|)(m+|y|)-nm$.

Correctness

Any valid final rectangle gives some shift $(x,y)$ between the original block and its rotated copy. Every place where those two blocks overlap is an old-old forced comparison, so the letters must match. Therefore our hash test accepts every shift that can come from a valid construction, and the construction must pay at least $|x|$ extra rows and $|y|$ extra columns.

Conversely, if our hash test accepts a shift, all old-old forced comparisons are consistent. Every old-new rotated pair can be filled with the needed letter, and every new-new pair can be filled arbitrarily but symmetrically. So a valid final rectangle exists with exactly $(n+|x|)(m+|y|)-nm$ invited dancers.

Thus minimizing over all accepted shifts gives the true minimum.

Complexity

Hash construction is $O(nm)$. There are fewer than $4nm$ shifts, each checked in $O(1)$. Memory usage is $O(nm)$.