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.
#include <bits/stdc++.h>
using namespace std;
inline int rint() {
  int a;
  scanf("%d", &a);
  return a;
}
inline long long rll() {
  long long a;
  scanf("%lld", &a);
  return a;
}
inline double rdouble() {
  double a;
  scanf("%lf", &a);
  return a;
}
int gcd(int a, int b) { return (b) ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
long long pow(long long a, long long b, long long q) {
  long long k = a;
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * k % q;
    k = k * k % q;
    b >>= 1;
  }
  return ans;
}
int main() {
  int n = rint(), m = rint();
  int a[105][105];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) a[i][j] = rint();
  vector<pair<string, int> > res;
  if (n <= m) {
    for (int i = 1; i <= n; i++) {
      int minn = a[i][1];
      for (int j = 1; j <= m; j++) minn = min(minn, a[i][j]);
      for (int j = 1; j <= minn; j++) res.push_back(make_pair("row", i));
      for (int j = 1; j <= m; j++) a[i][j] -= minn;
    }
    for (int i = 1; i <= m; i++) {
      int minn = a[1][i];
      for (int j = 1; j <= n; j++) minn = min(minn, a[j][i]);
      for (int j = 1; j <= minn; j++) res.push_back(make_pair("col", i));
      for (int j = 1; j <= n; j++) a[j][i] -= minn;
    }
  } else {
    for (int i = 1; i <= m; i++) {
      int minn = a[1][i];
      for (int j = 1; j <= n; j++) minn = min(minn, a[j][i]);
      for (int j = 1; j <= minn; j++) res.push_back(make_pair("col", i));
      for (int j = 1; j <= n; j++) a[j][i] -= minn;
    }
    for (int i = 1; i <= n; i++) {
      int minn = a[i][1];
      for (int j = 1; j <= m; j++) minn = min(minn, a[i][j]);
      for (int j = 1; j <= minn; j++) res.push_back(make_pair("row", i));
      for (int j = 1; j <= m; j++) a[i][j] -= minn;
    }
  }
  int all = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) all += a[i][j];
  if (all != 0) return puts("-1");
  cout << res.size() << endl;
  for (int i = 0; i < res.size(); i++)
    cout << res[i].first << ' ' << res[i].second << endl;
  return 0;
}
