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:

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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 inf = 1e9;
long long big_inf = 1e18;
int g[100][100];
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> g[i][j];
    }
  }
  vector<int> mn1(100, inf), mn2(100, inf);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      mn1[i] = min(mn1[i], g[i][j]);
      mn2[j] = min(mn2[j], g[i][j]);
    }
  }
  int ans = inf;
  vector<pair<int, int> > ans1;
  for (int i = 0; i <= mn1[0]; ++i) {
    bool can = true;
    int a[100], b[100];
    a[0] = i;
    for (int j = 0; j < m; ++j) {
      b[j] = g[0][j] - i;
    }
    if (!can) {
      continue;
    }
    for (int j = 1; j < n; ++j) {
      a[j] = g[j][0] - b[0];
      if (a[j] < 0) {
        can = false;
        break;
      }
    }
    if (!can) {
      continue;
    }
    for (int ii = 0; ii < n; ++ii) {
      for (int jj = 0; jj < m; ++jj) {
        if (a[ii] + b[jj] != g[ii][jj]) {
          can = false;
        }
      }
    }
    if (!can) {
      continue;
    }
    int new_ans = 0;
    for (int j = 0; j < n; ++j) {
      new_ans += a[j];
    }
    for (int j = 0; j < m; ++j) {
      new_ans += b[j];
    }
    if (new_ans < ans) {
      ans = new_ans;
      ans1 = {};
      for (int ii = 0; ii < n; ++ii) {
        for (int jj = 0; jj < a[ii]; ++jj) {
          ans1.push_back({1, ii + 1});
        }
      }
      for (int ii = 0; ii < m; ++ii) {
        for (int jj = 0; jj < b[ii]; ++jj) {
          ans1.push_back({2, ii + 1});
        }
      }
    }
  }
  if (ans == inf) {
    cout << -1;
  } else {
    cout << ans << "\n";
    for (int i = 0; i < ans1.size(); ++i) {
      if (ans1[i].first == 1) {
        cout << "row ";
      } else {
        cout << "col ";
      }
      cout << ans1[i].second << "\n";
    }
  }
  return 0;
}