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 n, m;
  cin >> n >> m;
  int a[n + 1][m + 1];
  int r[n + 1], c[m + 1], i, j, k, cnt = 0;
  int maxr = -1, minr = 501, maxj, minj, t = 0;
  bool f = false;
  memset(r, 0, sizeof(r));
  memset(c, 0, sizeof(c));
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      cin >> k;
      a[i][j] = k;
      if (k == 1) cnt++;
    }
  if (cnt == n * m) {
    if (n < m) {
      cout << n << endl;
      for (i = 1; i <= n; i++) cout << "row " << i << endl;
    } else {
      cout << m << endl;
      for (i = 1; i <= m; i++) cout << "col " << i << endl;
    }
    return 0;
  }
  if (n > m) {
    for (i = 1; i <= m; i++) {
      f = false;
      maxr = -1;
      minr = 501;
      for (j = 1; j <= n; j++) {
        if (a[j][i] > maxr) {
          maxr = a[j][i];
          maxj = i;
        }
        if (a[j][i] < minr) {
          minr = a[j][i];
          minj = i;
        }
      }
      if (minr == maxr) {
        if (minr == 0) {
          c[i] = 0;
        } else {
          c[i] = minr;
          f = true;
        }
      } else if (minr == 0)
        c[i] = 0;
      else {
        c[i] = minr;
        f = true;
      }
      if (f) {
        for (j = 1; j <= n; j++) {
          a[j][i] -= minr;
        }
      }
    }
    for (i = 1; i <= n; i++) {
      f = false;
      maxr = -1;
      minr = 501;
      for (j = 1; j <= m; j++) {
        if (a[i][j] > maxr) {
          maxr = a[i][j];
          maxj = j;
        }
        if (a[i][j] < minr) {
          minr = a[i][j];
          minj = j;
        }
      }
      if (minr == maxr) {
        if (minr == 0) {
          r[i] = 0;
        } else {
          r[i] = minr;
          f = true;
        }
      } else if (minr == 0)
        r[i] = 0;
      else {
        r[i] = minr;
        f = true;
      }
      if (f) {
        for (j = 1; j <= m; j++) {
          a[i][j] -= minr;
        }
      }
    }
  } else {
    for (i = 1; i <= n; i++) {
      f = false;
      maxr = -1;
      minr = 501;
      for (j = 1; j <= m; j++) {
        if (a[i][j] > maxr) {
          maxr = a[i][j];
          maxj = j;
        }
        if (a[i][j] < minr) {
          minr = a[i][j];
          minj = j;
        }
      }
      if (minr == maxr) {
        if (minr == 0) {
          r[i] = 0;
        } else {
          r[i] = minr;
          f = true;
        }
      } else if (minr == 0)
        r[i] = 0;
      else {
        r[i] = minr;
        f = true;
      }
      if (f) {
        for (j = 1; j <= m; j++) {
          a[i][j] -= minr;
        }
      }
    }
    for (i = 1; i <= m; i++) {
      f = false;
      maxr = -1;
      minr = 501;
      for (j = 1; j <= n; j++) {
        if (a[j][i] > maxr) {
          maxr = a[j][i];
          maxj = i;
        }
        if (a[j][i] < minr) {
          minr = a[j][i];
          minj = i;
        }
      }
      if (minr == maxr) {
        if (minr == 0) {
          c[i] = 0;
        } else {
          c[i] = minr;
          f = true;
        }
      } else if (minr == 0)
        c[i] = 0;
      else {
        c[i] = minr;
        f = true;
      }
      if (f) {
        for (j = 1; j <= n; j++) {
          a[j][i] -= minr;
        }
      }
    }
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++)
      if (a[i][j] != 0) {
        cout << "-1";
        return 0;
      }
  }
  for (i = 1; i <= n; i++) t += r[i];
  for (i = 1; i <= m; i++) t += c[i];
  cout << t << endl;
  for (i = 1; i <= n; i++)
    if (r[i] != 0) {
      while (r[i] != 0) {
        cout << "row " << i << endl;
        r[i] -= 1;
      }
    }
  for (i = 1; i <= m; i++)
    if (c[i] != 0) {
      while (c[i] != 0) {
        cout << "col " << i << endl;
        c[i] -= 1;
      }
    }
  return 0;
}