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 main() {
  int row, col;
  scanf("%d %d", &row, &col);
  vector<vector<int> > grid(row);
  for (vector<vector<int> >::iterator it = grid.begin(); it != grid.end();
       it++) {
    *it = vector<int>(col);
    for (vector<int>::iterator it2 = it->begin(); it2 != it->end(); it2++) {
      scanf("%d", &(*it2));
    }
  }
  int res = 0;
  vector<pair<string, int> > how;
  if (row <= col) {
    for (int r = 0; r < row; r++) {
      int minv = 987654321;
      for (int c = 0; c < col; c++) {
        minv = min(minv, grid[r][c]);
      }
      if (minv > 0) {
        for (int i = 0; i < minv; i++) {
          how.push_back(make_pair("row", r));
        }
        for (int c = 0; c < col; c++) {
          grid[r][c] -= minv;
        }
        res += minv;
      }
    }
    for (int c = 0; c < col; c++) {
      int minv = 987654321;
      for (int r = 0; r < row; r++) {
        minv = min(minv, grid[r][c]);
      }
      if (minv > 0) {
        for (int i = 0; i < minv; i++) {
          how.push_back(make_pair("col", c));
        }
        for (int r = 0; r < row; r++) {
          grid[r][c] -= minv;
        }
        res += minv;
      }
    }
  } else {
    for (int c = 0; c < col; c++) {
      int minv = 987654321;
      for (int r = 0; r < row; r++) {
        minv = min(minv, grid[r][c]);
      }
      if (minv > 0) {
        for (int i = 0; i < minv; i++) {
          how.push_back(make_pair("col", c));
        }
        for (int r = 0; r < row; r++) {
          grid[r][c] -= minv;
        }
        res += minv;
      }
    }
    for (int r = 0; r < row; r++) {
      int minv = 987654321;
      for (int c = 0; c < col; c++) {
        minv = min(minv, grid[r][c]);
      }
      if (minv > 0) {
        for (int i = 0; i < minv; i++) {
          how.push_back(make_pair("row", r));
        }
        for (int c = 0; c < col; c++) {
          grid[r][c] -= minv;
        }
        res += minv;
      }
    }
  }
  bool is_complete = true;
  for (int r = 0; r < row; r++) {
    for (int c = 0; c < col; c++) {
      if (grid[r][c] != 0) {
        is_complete = false;
        break;
      }
    }
  }
  if (is_complete) {
    printf("%d\n", res);
    for (vector<pair<string, int> >::iterator it = how.begin(); it != how.end();
         it++) {
      printf("%s %d\n", it->first.c_str(), it->second + 1);
    }
  } else {
    printf("-1\n");
  }
  return 0;
}