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:

<image>

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.
#!/usr/bin/env python3
import sys
from functools import reduce
from queue import PriorityQueue

INFINITY = 1000

def do_row(mat, actions):
    n = len(mat)
    did_nothing = True
    for r in range(n):
        mn, mx = min_and_max(mat[r])
        #print(mn, mx, file=sys.stderr)
        if mn > 0:
            dec = mn
            perform(mat, 'row', r, dec)
            for __ in range(dec):
                actions.append(('row', r + 1))
            did_nothing = False
    return did_nothing

def do_col(mat, actions):
    did_nothing = True
    for c, tup in enumerate(zip(*mat)):
        mn, mx = min_and_max(tup)
        if mn > 0:
            dec = mn
            perform(mat, 'col', c, dec)
            for __ in range(dec):
                actions.append(('col', c + 1))
            did_nothing = False
    return did_nothing


def main():
    n, m = map(int, sys.stdin.readline().split())
    mat = [None] * n
    actions = []
    for indx in range(n):
        mat[indx] = list(map(int, sys.stdin.readline().split()))

    while True:
        did_nothing = True
        if n <= m:
            did_nothing &= do_row(mat, actions)
            #print(mat, file=sys.stderr)
            did_nothing &= do_col(mat, actions)
            #print(mat, file=sys.stderr)
        else:
            did_nothing &= do_col(mat, actions)
            #print(mat, file=sys.stderr)
            did_nothing &= do_row(mat, actions)
            #print(mat, file=sys.stderr)

        if did_nothing:
            if max(map(max, mat)) == 0:
                print(len(actions))
                for act in actions:
                    print(' '.join(map(str, act)))
                return
            else:
                print(-1)
                return


def min_and_max(ary):
    return reduce(lambda m_M, cur: (min(m_M[0], cur), max(m_M[1], cur)), ary, (INFINITY, -1))

def perform(mat, r_or_c, _indx, dec):
    if r_or_c == 'row':
        for indx in range(len(mat[_indx])):
            mat[_indx][indx] -= dec
    else:
        for indx in range(len(mat)):
            mat[indx][_indx] -= dec



if __name__ == '__main__':
    main()
