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>
#pragma GCC optimize("O2")
using namespace std;
long long n, m;
int a[102][102];
int maxl[102], minl[102];
int maxc[102], minc[102];
vector<pair<int, int> > v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      cin >> a[i][j];
      if (j == 1)
        minl[i] = maxl[i] = a[i][j];
      else
        minl[i] = min(minl[i], a[i][j]), maxl[i] = max(maxl[i], a[i][j]);
    }
  for (int j = 1; j <= m; ++j)
    for (int i = 1; i <= n; ++i) {
      if (i == 1)
        minc[j] = maxc[j] = a[i][j];
      else
        minc[j] = min(minc[j], a[i][j]), maxc[i] = max(maxc[j], a[i][j]);
    }
  if (n >= m) {
    for (int i = 1; i <= n; ++i) {
      int minx = minl[i];
      for (int j = 1; j <= m; ++j) minx = min(minx, a[i][j] - minc[j]);
      for (int q = 1; q <= minx; ++q) v.push_back({1, i});
      for (int j = 1; j <= m; ++j) a[i][j] -= minx;
    }
    for (int j = 1; j <= m; ++j) {
      int maxi = 0;
      int mini = 1e9;
      for (int i = 1; i <= n; ++i) {
        maxi = max(maxi, a[i][j]);
        mini = min(mini, a[i][j]);
      }
      if (maxi == mini) {
        for (int i = 1; i <= n; ++i) a[i][j] = 0;
        for (int q = 1; q <= maxi; ++q) v.push_back({2, j});
      } else {
        cout << -1;
        return 0;
      }
    }
  } else {
    for (int i = 1; i <= m; ++i) {
      int minx = minc[i];
      for (int j = 1; j <= n; ++j) minx = min(minx, a[j][i] - minl[j]);
      for (int q = 1; q <= minx; ++q) v.push_back({2, i});
      for (int j = 1; j <= n; ++j) a[j][i] -= minx;
    }
    for (int j = 1; j <= n; ++j) {
      int maxi = 0;
      int mini = 1e9;
      for (int i = 1; i <= m; ++i) {
        maxi = max(maxi, a[j][i]);
        mini = min(mini, a[j][i]);
      }
      if (maxi == mini) {
        for (int i = 1; i <= m; ++i) a[j][i] = 0;
        for (int q = 1; q <= maxi; ++q) v.push_back({1, j});
      } else {
        cout << -1;
        return 0;
      }
    }
  }
  cout << v.size() << '\n';
  for (int i = 0; i < v.size(); ++i) {
    if (v[i].first == 1)
      cout << "row ";
    else
      cout << "col ";
    cout << v[i].second << '\n';
  }
  return 0;
}
