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;
int a[100][100], b[100][100];
int n, m;
bool comp(int row, int col) {
  if (row == -1) {
    for (int i = 0; i < n; i++)
      if (b[i][col] <= a[i][col])
        continue;
      else
        return false;
    return true;
  } else {
    for (int j = 0; j < m; j++)
      if (b[row][j] <= a[row][j])
        continue;
      else
        return false;
    return true;
  }
}
void inc(int row, int col) {
  if (row == -1)
    for (int i = 0; i < n; i++) b[i][col]++;
  else
    for (int j = 0; j < m; j++) b[row][j]++;
  return;
}
void dec(int row, int col) {
  if (row == -1)
    for (int i = 0; i < n; i++) b[i][col]--;
  else
    for (int j = 0; j < m; j++) b[row][j]--;
  return;
}
bool compare() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (a[i][j] == b[i][j])
        continue;
      else
        return false;
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  string sol = "";
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> a[i][j], b[i][j] = 0;
  int tot_count = 0;
  if (n < m) {
    for (int i = 0; i < n; i++) {
      string s = "row ", app = to_string(i + 1);
      s += app;
      int count = 0;
      while (comp(i, -1)) inc(i, -1), count++;
      dec(i, -1), count--;
      while (count--) tot_count++, sol.append(s), sol += '\n';
    }
    for (int j = 0; j < m; j++) {
      string s = "col ", app = to_string(j + 1);
      s += app;
      int count = 0;
      while (comp(-1, j)) inc(-1, j), count++;
      dec(-1, j), count--;
      while (count--) tot_count++, sol.append(s), sol += '\n';
    }
  } else {
    for (int j = 0; j < m; j++) {
      string s = "col ", app = to_string(j + 1);
      s += app;
      int count = 0;
      while (comp(-1, j)) inc(-1, j), count++;
      dec(-1, j), count--;
      while (count--) tot_count++, sol.append(s), sol += '\n';
    }
    for (int i = 0; i < n; i++) {
      string s = "row ", app = to_string(i + 1);
      s += app;
      int count = 0;
      while (comp(i, -1)) inc(i, -1), count++;
      dec(i, -1), count--;
      while (count--) tot_count++, sol.append(s), sol += '\n';
    }
  }
  if (compare())
    cout << tot_count << endl << sol << endl;
  else
    cout << -1 << endl;
  return 0;
}