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;
using lli = long long int;
using pii = pair<int, int>;
using vi = vector<int>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vlli = vector<long long int>;
using vpii = vector<pair<int, int>>;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vi row(n), col(m);
  vvi grid(n, vi(m)), level(n, vi(m));
  vpii ans1, ans2, ans;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> grid[i][j];
      level[i][j] = grid[i][j];
    }
  for (int i = 0; i < n; i++) {
    int x = 0x3f3f3f3f;
    for (int j = 0; j < m; j++) x = min(x, grid[i][j]);
    row[i] = x;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) grid[i][j] -= row[i];
    for (int j = 0; j < row[i]; j++) ans1.push_back(make_pair(0, i + 1));
  }
  for (int i = 0; i < m; i++) {
    int x = 0x3f3f3f3f;
    for (int j = 0; j < n; j++) x = min(x, grid[j][i]);
    col[i] = x;
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) grid[j][i] -= col[i];
    for (int j = 0; j < col[i]; j++) ans1.push_back(make_pair(1, i + 1));
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (grid[i][j] != 0) {
        cout << -1;
        return 0;
      }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) grid[i][j] = level[i][j];
  for (int i = 0; i < m; i++) {
    int x = 0x3f3f3f3f;
    for (int j = 0; j < n; j++) x = min(x, grid[j][i]);
    col[i] = x;
  }
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) grid[j][i] -= col[i];
    for (int j = 0; j < col[i]; j++) ans2.push_back(make_pair(1, i + 1));
  }
  for (int i = 0; i < n; i++) {
    int x = 0x3f3f3f3f;
    for (int j = 0; j < m; j++) x = min(x, grid[i][j]);
    row[i] = x;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) grid[i][j] -= row[i];
    for (int j = 0; j < row[i]; j++) ans2.push_back(make_pair(0, i + 1));
  }
  if (ans1.size() < ans2.size())
    ans = ans1;
  else
    ans = ans2;
  cout << ans.size() << '\n';
  for (int i = 0; i < ans.size(); i++)
    if (ans[i].first == 0)
      cout << "row " << ans[i].second << '\n';
    else
      cout << "col " << ans[i].second << '\n';
}