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 long long PINF = 1E9;
const long long NINF = -1E9;
const long long M = 1E9 + 7;
const double EPS = 1E-9;
void print(pair<long long, pair<vector<long long>, vector<long long> > >& x) {
  long long ans = x.first;
  cout << ans << "\n";
  vector<long long>& row = x.second.first;
  vector<long long>& col = x.second.second;
  int n = (int)row.size();
  int m = (int)col.size();
  for (int i = 0; i < n; i++) {
    for (int k = 0; k < row[i]; k++) {
      cout << "row " << i + 1 << "\n";
    }
  }
  for (int j = 0; j < m; j++) {
    for (int k = 0; k < col[j]; k++) {
      cout << "col " << j + 1 << "\n";
    }
  }
}
pair<long long, pair<vector<long long>, vector<long long> > > f(
    vector<vector<long long> > g) {
  int n = g.size();
  int m = g[0].size();
  vector<long long> row(n), col(m);
  for (int i = 0; i < n; i++) {
    row[i] = *min_element(g[i].begin(), g[i].end());
    for (int j = 0; j < m; j++) {
      g[i][j] -= row[i];
    }
  }
  bool f1 = true, f2 = true;
  for (int j = 0; j < m; j++) {
    for (int i = 0; i < n; i++) {
      if (i > 0 && g[i][j] != g[i - 1][j]) {
        f2 = false;
        break;
      }
    }
    if (!f2) {
      f1 = false;
      break;
    }
    col[j] = g[0][j];
  }
  long long ans = PINF;
  if (f1) {
    ans = 0;
    for (int i = 0; i < n; i++) {
      ans += row[i];
    }
    for (int j = 0; j < m; j++) {
      ans += col[j];
    }
  }
  return make_pair(ans, make_pair(row, col));
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<long long> > g(n, vector<long long>(m)),
      h(m, vector<long long>(n));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> g[i][j];
    }
  }
  for (int j = 0; j < m; j++) {
    for (int i = 0; i < n; i++) {
      h[j][i] = g[i][j];
    }
  }
  auto x1 = f(g);
  auto x2 = f(h);
  if (x1.first == PINF && x2.first == PINF) {
    cout << -1 << "\n";
  } else {
    if (x1.first < x2.first) {
      print(x1);
    } else {
      swap(x2.second.first, x2.second.second);
      print(x2);
    }
  }
  return 0;
}