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 int inf = INT_MAX;
const int maxn = 107;
int n, m, a[maxn][maxn], row[maxn], col[maxn];
struct NODE {
  int index;
  char rc;
  NODE() {}
  NODE(int idx, char _) { index = idx, rc = _; }
  friend bool operator<(NODE a, NODE b) {
    int al, bl;
    if (a.rc == 'r') {
      al = row[a.index];
    } else {
      al = col[a.index];
    }
    if (b.rc == 'r') {
      bl = row[b.index];
    } else {
      bl = col[b.index];
    }
    if (al == bl) {
      if (n >= m) {
        if (a.rc == 'r' && b.rc == 'c') {
          return true;
        } else
          return false;
      } else {
        if (a.rc == 'c' && b.rc == 'r') {
          return true;
        } else
          return false;
      }
    }
    return al > bl;
  }
};
string num2str(int i) {
  stringstream ss;
  ss << i;
  return ss.str();
}
int main() {
  while (cin >> n >> m) {
    vector<string> ans;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        cin >> a[i][j];
      }
    }
    priority_queue<NODE> q;
    do {
      if (!q.empty()) {
        NODE top = q.top();
        q.pop();
        if (top.rc == 'r') {
          int i = top.index;
          int val = row[i];
          for (int j = 1; j <= m; j++) {
            a[i][j] -= val;
          }
          while (val--) ans.push_back("row " + num2str(i));
        } else {
          int j = top.index;
          int val = col[j];
          for (int i = 1; i <= n; i++) {
            a[i][j] -= val;
          }
          while (val--) ans.push_back("col " + num2str(j));
        }
      }
      while (!q.empty()) q.pop();
      for (int i = 1; i <= n; i++) {
        row[i] = inf;
        for (int j = 1; j <= m; j++) {
          if (a[i][j] < row[i]) row[i] = a[i][j];
        }
        if (row[i]) {
          q.push(NODE(i, 'r'));
        }
      }
      for (int j = 1; j <= m; j++) {
        col[j] = inf;
        for (int i = 1; i <= n; i++) {
          if (a[i][j] < col[j]) col[j] = a[i][j];
        }
        if (col[j]) {
          q.push(NODE(j, 'c'));
        }
      }
    } while (!q.empty());
    int flag = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        if (a[i][j]) flag = 1;
      }
    }
    if (flag) {
      cout << -1 << endl;
    } else {
      int len = ans.size();
      cout << len << endl;
      for (int i = 0; i < len; i++) {
        cout << ans[i] << endl;
      }
    }
  }
  return 0;
}