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>
using namespace std;
const int N = 110;
vector<int> row, col, row2, col2;
int mat[N][N], mat2[N][N];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> mat[i][j];
      mat2[i][j] = mat[i][j];
    }
  while (true) {
    bool all_zero = true;
    bool get = false;
    for (int i = 0; i < n; i++) {
      int min_elem = 501;
      for (int j = 0; j < m; j++) min_elem = min(mat[i][j], min_elem);
      if (min_elem != 0) {
        for (int t = 0; t < min_elem; t++) row.push_back(i);
        get = true;
      }
      for (int j = 0; j < m; j++) mat[i][j] -= min_elem;
    }
    for (int j = 0; j < m; j++) {
      int min_elem = 501;
      for (int i = 0; i < n; i++) min_elem = min(mat[i][j], min_elem);
      if (min_elem != 0) {
        get = true;
        for (int t = 0; t < min_elem; t++) col.push_back(j);
      }
      for (int i = 0; i < n; i++) mat[i][j] -= min_elem;
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (mat[i][j] != 0) all_zero = false;
    if (!get && !all_zero) {
      cout << -1 << endl;
      return 0;
    }
    if (all_zero) {
      break;
    }
  }
  while (true) {
    bool all_zero = true;
    for (int j = 0; j < m; j++) {
      int min_elem = 501;
      for (int i = 0; i < n; i++) min_elem = min(mat2[i][j], min_elem);
      if (min_elem != 0) {
        for (int t = 0; t < min_elem; t++) col2.push_back(j);
      }
      for (int i = 0; i < n; i++) mat2[i][j] -= min_elem;
    }
    for (int i = 0; i < n; i++) {
      int min_elem = 501;
      for (int j = 0; j < m; j++) min_elem = min(mat2[i][j], min_elem);
      if (min_elem != 0) {
        for (int t = 0; t < min_elem; t++) row2.push_back(i);
      }
      for (int j = 0; j < m; j++) mat2[i][j] -= min_elem;
    }
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++)
        if (mat2[i][j] != 0) all_zero = false;
    if (all_zero) {
      break;
    }
  }
  if (col.size() + row.size() <= col2.size() + row2.size()) {
    cout << row.size() + col.size() << endl;
    for (int i = 0; i < row.size(); i++) cout << "row " << row[i] + 1 << endl;
    for (int i = 0; i < col.size(); i++) cout << "col " << col[i] + 1 << endl;
  } else {
    cout << row2.size() + col2.size() << endl;
    for (int i = 0; i < row2.size(); i++) cout << "row " << row2[i] + 1 << endl;
    for (int i = 0; i < col2.size(); i++) cout << "col " << col2[i] + 1 << endl;
  }
}