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;
const long long N = 1000006, MOD = 1e9 + 7, INF = 1e16;
long long a[N], b, n, m, k, g[105][105], q, x, ans, Ans, pre[N], l, r;
string s;
pair<int, int> p[N];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> g[i][j];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (g[i][j] < 1) x = 1;
  while (!x) {
    Ans++;
    q += min(n, m);
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) {
        g[i][j]--;
        if (g[i][j] < 1) x = 1;
      }
  }
  ans = 0;
  x = 0;
  for (int i = 1; i <= n; i++)
    if (!g[i][1]) x = 1;
  while (!x) {
    ans++, r++;
    for (int i = 1; i <= n; i++) {
      g[i][1]--;
      if (!g[i][1]) x = 1;
    }
  }
  x = 0;
  for (int i = 1; i <= m; i++)
    if (!g[1][i]) x = 1;
  while (!x) {
    ans++, l++;
    for (int i = 1; i <= m; i++) {
      g[1][i]--;
      if (!g[1][i]) x = 1;
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (g[i][j] != g[1][j] + g[i][1]) {
        cout << -1;
        return 0;
      }
  for (int i = 1; i <= n; i++) ans += g[i][1];
  for (int i = 2; i <= m; i++) ans += g[1][i];
  cout << ans + q << endl;
  for (int i = 1; i <= l; i++)
    cout << "row"
         << " " << 1 << endl;
  for (int i = 1; i <= r; i++)
    cout << "col"
         << " " << 1 << endl;
  for (int i = 1; i <= n; i++)
    for (int u = 1; u <= g[i][1]; u++)
      cout << "row"
           << " " << i << endl;
  for (int i = 1; i <= m; i++)
    for (int u = 1; u <= g[1][i]; u++)
      cout << "col"
           << " " << i << endl;
  if (n < m)
    s = "row";
  else
    s = "col";
  for (int i = 0; i < min(n, m) * Ans; i++)
    cout << s << " " << i % min(n, m) + 1 << endl;
}