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Problem A - Snakey String With Solution
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A **snakey string** is a fancy rendering of an otherwise normal string of text. When a string is made snakey, it is placed on a 2D grid such that the following conditions are met:

* The first character of the string is on the first column of the grid, the second on the second column, …, and the last on the last column.
* Each character in the string occupies either the row directly above or directly below the row of the previous character.

::

      G
    I   S   U
  B       O   T
                H

*Figure 1: The snakey string in the sample case.*

Given a snakey string, can you recover the original string?

Input
=====

The first line of input contains two integers **r** and **c** (2 ≤ r, c ≤ 100), the number of rows and columns of the grid.

The next **r** lines each contain **c** characters that form the 2D grid containing the snakey string.
Empty cells are encoded with a period (`.`), while uppercase letters (A–Z) represent characters in the original string.
Every column in the grid contains **exactly one** uppercase letter. It is possible that some rows contain no uppercase letters.

Output
======

Output a single string — the original string that was used to build this snakey string.


Sample Input 1
==============

::

  4 8
  ..G.....
  .I.S.U..
  B...O.T.
  .......H

Sample Output 1
===============

::

  BIGSOUTH


Here is a **simple explanation** and a **clean Python 3 solution**.

Solution
========

Explanation
-----------

* You are given an **r × c** grid.
* **Each column contains exactly ONE uppercase letter** (the others are dots).
* To recover the original string, you just:

  1. Look at **each column from left to right**.
  2. Find the **uppercase letter** in that column.
  3. Append it to the result.

That's it—no actual snake traversal is needed because the puzzle guarantees exactly one letter per column.

Python 3 Solution
-----------------

.. code-block:: python3

  def solve():
      r, c = map(int, input().split())
      grid = [input().strip() for _ in range(r)]

      result = []

      # For each column, find the one uppercase letter
      for col in range(c):
          for row in range(r):
              ch = grid[row][col]
              if 'A' <= ch <= 'Z':  # meaningful letter
                  result.append(ch)
                  break

      print("".join(result))

  if __name__ == "__main__":
      solve()