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 maxn = 500 + 13;
const int Max = 0x7ffffffa;
int Grid[maxn][maxn];
int n, m;
struct Node {
  string name;
  int id;
  Node() {}
  Node(string _name, int _id) {
    name = _name;
    id = _id;
  }
  bool operator<(const Node t) const { return id > t.id; }
};
int GetMinRow(int nt) {
  int MinNumber = Max;
  for (int i = 0; i < m; ++i) {
    MinNumber = min(MinNumber, Grid[nt][i]);
  }
  return MinNumber;
}
int GetMinCol(int nt) {
  int MinNumber = Max;
  for (int i = 0; i < n; ++i) {
    MinNumber = min(MinNumber, Grid[i][nt]);
  }
  return MinNumber;
}
void SolveRow(stack<Node>& tmp) {
  int moveNumber = 0;
  for (int i = 0; i < n; ++i) {
    int minNumber = GetMinRow(i);
    if (minNumber) {
      for (int j = 0; j < minNumber; ++j) tmp.push(Node("row", i + 1));
      for (int j = 0; j < m; ++j) {
        Grid[i][j] -= minNumber;
      }
    }
  }
}
void SolveCol(stack<Node>& tmp) {
  for (int i = 0; i < m; ++i) {
    int minNumber = GetMinCol(i);
    if (minNumber) {
      for (int j = 0; j < minNumber; ++j) {
        tmp.push(Node("col", i + 1));
      }
      for (int j = 0; j < n; ++j) {
        Grid[j][i] -= minNumber;
      }
    }
  }
}
bool JugIsEmpty() {
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j)
      if (Grid[i][j]) return false;
  return true;
}
stack<Node> Solve() {
  stack<Node> res;
  int moveNumber = 0;
  if (n < m) {
    SolveRow(res);
    SolveCol(res);
  } else {
    SolveCol(res);
    SolveRow(res);
  }
  return res;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < m; ++j) {
      scanf("%d", &Grid[i][j]);
    }
  stack<Node> res = Solve();
  if (JugIsEmpty()) {
    priority_queue<Node> Line;
    cout << res.size() << "\n";
    while (!res.empty()) {
      Node it = res.top();
      res.pop();
      Line.push(it);
    }
    while (!Line.empty()) {
      Node it = Line.top();
      Line.pop();
      cout << it.name << " " << it.id << "\n";
    }
  } else {
    cout << "-1\n";
  }
  return 0;
}