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;
const int inf = 1e9;
struct mve {
  int i, t, w;
  mve(int a, int b, int c) : i(a), t(b), w(c) {}
  mve() {}
};
int n, m, cant;
int minx[110], miny[110], mx[110];
int grid[110][110];
string s[2] = {"row", "col"};
vector<mve> moves;
void solve_row() {
  for (int i = 0; i < (int)n; i++)
    if (minx[i]) {
      moves.push_back(mve(i, 0, minx[i]));
      cant += minx[i];
      for (int j = 0; j < (int)m; j++) grid[i][j] -= minx[i];
    }
}
void solve_col() {
  for (int j = 0; j < (int)m; j++)
    if (miny[j]) {
      moves.push_back(mve(j, 1, miny[j]));
      cant += miny[j];
      for (int i = 0; i < (int)n; i++) grid[i][j] -= miny[j];
    }
}
bool solve_row2() {
  for (int i = 0; i < (int)n; i++)
    for (int j = 0; j < (int)m; j++) {
      minx[i] = min(minx[i], grid[i][j]);
      mx[i] = max(mx[i], grid[i][j]);
    }
  for (int i = 0; i < (int)n; i++) {
    if (minx[i]) {
      moves.push_back(mve(i, 0, minx[i]));
      cant += minx[i];
    }
    if (minx[i] < mx[i]) {
      printf("-1\n");
      return 0;
    }
  }
  return 1;
}
bool solve_col2() {
  for (int i = 0; i < (int)n; i++)
    for (int j = 0; j < (int)m; j++) {
      miny[j] = min(miny[j], grid[i][j]);
      mx[j] = max(mx[j], grid[i][j]);
    }
  for (int j = 0; j < (int)m; j++) {
    if (miny[j]) {
      moves.push_back(mve(j, 1, miny[j]));
      cant += miny[j];
    }
    if (miny[j] < mx[j]) {
      printf("-1\n");
      return 0;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 0; i < (int)110; i++) {
    minx[i] = inf;
    miny[i] = inf;
    mx[i] = 0;
  }
  for (int i = 0; i < (int)n; i++)
    for (int j = 0; j < (int)m; j++) {
      scanf("%d", &grid[i][j]);
      minx[i] = min(minx[i], grid[i][j]);
      miny[j] = min(miny[j], grid[i][j]);
    }
  cant = 0;
  if (n > m) {
    solve_col();
    if (!solve_row2()) return 0;
  } else {
    solve_row();
    if (!solve_col2()) return 0;
  }
  printf("%d\n", cant);
  for (int i = 0; i < (int)moves.size(); i++) {
    while (moves[i].w--) {
      cout << s[moves[i].t];
      printf(" %d\n", moves[i].i + 1);
    }
  }
  return 0;
}