On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.
import sys

dimensions = [int(i) for i in raw_input().split()]

rows = dimensions[0]
cols = dimensions[1]

arr = []

for t in xrange(rows):
    stuff = [int(i) for i in raw_input().split()]
    arr.append(list(stuff))

row_increments = []
for i in xrange(rows):
    row_increments.append(0)

col_increments = []

for i in xrange(cols):
    col_increments.append(0)

min_top_row = min(arr[0])

#valid because there exists a solution where some column is never pressed
#also optimal because the transformations being made below are actually forced

for i in xrange(cols):
    col_increments[i] = arr[0][i] - min_top_row

for row in xrange(rows):
    for col in xrange(cols):
        arr[row][col] -= col_increments[col]

min_left_col = min([arr[i][0] for i in xrange(rows)])

for i in xrange(rows):
    row_increments[i] = arr[i][0] - min_left_col

for row in xrange(rows):
    for col in xrange(cols):
        arr[row][col] -= row_increments[row]

#by this stage, all elements of the array should be equal
#we choose to finish the job by spamming rows or columns depending on which takes less moves

if arr[0][0] < 0:
    print -1
    sys.exit()
    
for i in xrange(rows):
    for j in xrange(cols):
        if arr[i][j] != arr[0][0]:
            print -1
            sys.exit()

if rows < cols:
    for i in xrange(rows):
        row_increments[i] += arr[0][0]
else:
    for i in xrange(cols):
        col_increments[i] += arr[0][0]

print sum(row_increments) + sum(col_increments)

for i in xrange(rows):
    for j in xrange(row_increments[i]):
        print "row " + str(i+1)

for i in xrange(cols):
    for j in xrange(col_increments[i]):
        print "col " + str(i+1)
        

