Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int grid[100][100];
int main() {
  int rows, cols;
  scanf("%d %d", &rows, &cols);
  int rows_min[100];
  int cols_min[100];
  std::fill(rows_min, rows_min + rows, 500);
  std::fill(cols_min, cols_min + cols, 500);
  for (int i = 0; i < rows; ++i) {
    for (int j = 0; j < cols; ++j) {
      scanf("%d", &grid[i][j]);
      rows_min[i] = std::min(grid[i][j], rows_min[i]);
      cols_min[j] = std::min(grid[i][j], cols_min[j]);
    }
  }
  std::vector<std::string> v;
  if (cols >= rows) {
    for (int i = 0; i < rows; ++i) {
      for (int j = 0; j < rows_min[i]; ++j)
        v.emplace_back("row " + std::to_string(i + 1));
      for (int j = 0; j < cols; ++j) {
        grid[i][j] -= rows_min[i];
        cols_min[j] = std::min(grid[i][j], cols_min[j]);
      }
      rows_min[i] = 0;
    }
    for (int j = 0; j < cols; ++j) {
      for (int i = 0; i < cols_min[j]; ++i)
        v.emplace_back("col " + std::to_string(j + 1));
      for (int i = 0; i < rows; ++i) {
        grid[i][j] -= cols_min[j];
        rows_min[i] = std::min(grid[i][j], rows_min[i]);
      }
      cols_min[j] = 0;
    }
  } else {
    for (int j = 0; j < cols; ++j) {
      for (int i = 0; i < cols_min[j]; ++i)
        v.emplace_back("col " + std::to_string(j + 1));
      for (int i = 0; i < rows; ++i) {
        grid[i][j] -= cols_min[j];
        rows_min[i] = std::min(grid[i][j], rows_min[i]);
      }
      cols_min[j] = 0;
    }
    for (int i = 0; i < rows; ++i) {
      for (int j = 0; j < rows_min[i]; ++j)
        v.emplace_back("row " + std::to_string(i + 1));
      for (int j = 0; j < cols; ++j) {
        grid[i][j] -= rows_min[i];
        cols_min[j] = std::min(grid[i][j], cols_min[j]);
      }
      rows_min[i] = 0;
    }
  }
  bool ok = true;
  for (int i = 0; i < rows * cols; ++i) {
    if (grid[i / cols][i % cols]) {
      ok = false;
      break;
    }
  }
  if (!ok) {
    printf("-1\n");
  } else {
    printf("%zu\n", v.size());
    std::for_each(v.begin(), v.end(),
                  [](const std::string &s) { printf("%s\n", s.c_str()); });
  }
  return 0;
}