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;
class Solution {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int n, m;
    in >> n >> m;
    std::vector<std::vector<int>> a;
    a.resize(std::min(n, m));
    for (auto& b : a) b.resize(std::max(m, n));
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < m; ++j) {
        int val;
        in >> val;
        if (n < m)
          a[i][j] = val;
        else
          a[j][i] = val;
      }
    bool order = n < m;
    std::tie(n, m) = std::pair<int, int>(std::minmax(n, m));
    struct sol_el {
      int count;
      bool row;
      int index;
    };
    int step_cnt = 0;
    std::vector<sol_el> sol;
    for (int i = 0; i < n; ++i) {
      auto cnt = *std::min_element(a[i].begin(), a[i].end());
      if (cnt > 0) {
        step_cnt += cnt;
        sol.push_back({cnt, order, i});
        for (auto& b : a[i]) b -= cnt;
      }
    }
    for (int i = 0; i < m; ++i) {
      int cnt = std::numeric_limits<int>::max();
      for (int j = 0; j < n; ++j) cnt = std::min(a[j][i], cnt);
      if (cnt > 0) {
        step_cnt += cnt;
        sol.push_back({cnt, !order, i});
        for (int j = 0; j < n; ++j) a[j][i] -= cnt;
      }
    }
    int sum = 0;
    for (auto& b : a) sum += accumulate(b.begin(), b.end(), 0);
    if (sum == 0) {
      out << step_cnt << '\n';
      for (auto& s : sol) {
        for (int i = 0; i < s.count; ++i) {
          out << (s.row ? "row " : "col ");
          out << s.index + 1 << '\n';
        }
      }
    } else
      out << "-1\n";
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}