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:

<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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int find_min(const vector<int>& m, const vector<int>& ops) {
  int r = m.front();
  for (int i = 0; i < m.size(); ++i) {
    r = min(r, m[i] - ops[i]);
  }
  return r;
}
int main() {
  int n, m;
  std::cin >> n >> m;
  vector<vector<int>> nums(min(n, m), vector<int>(max(n, m), 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> (n <= m ? nums[i][j] : nums[j][i]);
    }
  }
  bool is_swap = false;
  if (n > m) {
    is_swap = true;
    swap(n, m);
  }
  int res = 0;
  vector<int> cols_ops(m, 0);
  vector<int> rows_ops(n, 0);
  int m_first = find_min(nums[0], cols_ops);
  for (int i = 0; i < nums[0].size(); ++i) {
    cols_ops[i] = nums[0][i] - m_first;
    res += cols_ops[i];
  }
  res += (rows_ops[0] = m_first);
  for (int r = 1; r < n; ++r) {
    int lowest = find_min(nums[r], cols_ops);
    for (int i = 0; i < nums[r].size(); ++i) {
      if (nums[r][i] - cols_ops[i] != lowest) {
        cout << -1 << endl;
        return 0;
      }
    }
    rows_ops[r] = lowest;
    res += lowest;
  }
  std::cout << res << std::endl;
  for (int i = 0; i < n; ++i) {
    for (int index = 0; index < rows_ops[i]; ++index) {
      cout << (is_swap ? "col " : "row ") << (i + 1) << "\n";
    }
  }
  for (int i = 0; i < m; ++i) {
    for (int index = 0; index < cols_ops[i]; ++index) {
      cout << (is_swap ? "row " : "col ") << (i + 1) << "\n";
    }
  }
  return 0;
}