You've got a 5 × 5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to bottom, let's index the matrix columns by numbers from 1 to 5 from left to right. In one move, you are allowed to apply one of the two following transformations to the matrix:
 1.  Swap two neighboring matrix rows, that is, rows with indexes *i* and *i* + 1 for some integer *i* (1 ≤ *i* &lt; 5). 1.  Swap two neighboring matrix columns, that is, columns with indexes *j* and *j* + 1 for some integer *j* (1 ≤ *j* &lt; 5). 
You think that a matrix looks beautiful, if the single number one of the matrix is located in its middle (in the cell that is on the intersection of the third row and the third column). Count the minimum number of moves needed to make the matrix beautiful.
The input consists of five lines, each line contains five integers: the *j*-th integer in the *i*-th line of the input represents the element of the matrix that is located on the intersection of the *i*-th row and the *j*-th column. It is guaranteed that the matrix consists of 24 zeroes and a single number one.
import numpy as np

n, m = map(int, input().split()) # taking number of rows and column
matrix = np.array([input().strip().split() for _ in range(n)], int)

  # Find the current row and column of the number one.
row_one, col_one = np.where(matrix == 1)[0][0], np.where(matrix == 1)[1][0]

  # If the number one is already in the middle row and column, then we are done and the minimum number of moves is zero.
if row_one == 2 and col_one == 2:
  print(0)

  # Otherwise, we need to swap rows and columns until the number one is in the middle.
num_moves = 0
if row_one < 2:
  num_moves += row_one
  matrix = (matrix, row_one, row_one + 1)
elif row_one > 2:
  num_moves += 4 - row_one
  matrix = (matrix, row_one, row_one - 1)

if col_one < 2:
  num_moves += col_one
  matrix = (matrix, col_one, col_one + 1)
elif col_one > 2:
  num_moves += 4 - col_one
  matrix = (matrix, col_one, col_one - 1)

print (num_moves)
