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;
void upd1(long long i, vector<vector<long long>> &mx, long long val) {
  for (int j = 0; j < mx[i].size(); ++j) mx[i][j] -= val;
}
void upd2(long long j, vector<vector<long long>> &mx, long long val) {
  for (int i = 0; i < mx.size(); ++i) {
    mx[i][j] -= val;
  }
}
int main() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> mx(n, vector<long long>(m));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> mx[i][j];
    }
  }
  vector<pair<pair<long long, long long>, long long>> ans;
  long long minn = 1e18;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      minn = min(minn, mx[i][j]);
    }
  }
  if (minn > 0) {
    if (n < m) {
      for (int i = 0; i < n; ++i) {
        ans.push_back({{1, minn}, i});
        upd1(i, mx, minn);
      }
    } else {
      for (int i = 0; i < m; ++i) {
        ans.push_back({{2, minn}, i});
        upd2(i, mx, minn);
      }
    }
  }
  if (n <= m) {
    long long cnt = 1;
    while (cnt != 0) {
      cnt = 0;
      long long maxx = 1e18;
      for (int i = 0; i < n; ++i) {
        maxx = 1e18;
        for (int j = 0; j < m; ++j) maxx = min(maxx, mx[i][j]);
        if (maxx > 0) {
          ++cnt;
          upd1(i, mx, maxx);
          ans.push_back({{1, maxx}, i});
        }
      }
      for (int j = 0; j < m; ++j) {
        maxx = 1e18;
        for (int i = 0; i < n; ++i) maxx = min(maxx, mx[i][j]);
        if (maxx > 0) {
          ++cnt;
          upd2(j, mx, maxx);
          ans.push_back({{2, maxx}, j});
        }
      }
    }
  } else {
    long long cnt = 1;
    while (cnt != 0) {
      cnt = 0;
      long long maxx = 1e18;
      for (int j = 0; j < m; ++j) {
        maxx = 1e18;
        for (int i = 0; i < n; ++i) maxx = min(maxx, mx[i][j]);
        if (maxx > 0) {
          ++cnt;
          upd2(j, mx, maxx);
          ans.push_back({{2, maxx}, j});
        }
      }
      for (int i = 0; i < n; ++i) {
        maxx = 1e18;
        for (int j = 0; j < m; ++j) maxx = min(maxx, mx[i][j]);
        if (maxx > 0) {
          ++cnt;
          upd1(i, mx, maxx);
          ans.push_back({{1, maxx}, i});
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (mx[i][j] != 0) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  long long kolvo = 0;
  for (int i = 0; i < ans.size(); ++i) {
    kolvo += ans[i].first.second;
  }
  cout << kolvo << endl;
  string s1 = "row", s2 = "col", s;
  for (int i = 0; i < ans.size(); ++i) {
    if (ans[i].first.first == 2)
      s = s2;
    else
      s = s1;
    for (int j = 0; j < ans[i].first.second; ++j) {
      cout << s << " " << ans[i].second + 1 << endl;
    }
  }
  return 0;
}
