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 ll = long long;
using TArray = std::vector<std::vector<ll>>;
struct TSolution {
  std::vector<ll> Rows;
  std::vector<ll> Cols;
  bool Inverted = false;
  bool IsGood = false;
  ll GetTotal() const { return Rows.size() + Cols.size(); }
  bool IsBetter(const TSolution& other) const {
    if (!IsGood) {
      return false;
    }
    if (GetTotal() == 0) {
      return false;
    }
    return GetTotal() < other.GetTotal();
  }
};
bool HasNonZeroes(const TArray& array) {
  for (const auto& r : array) {
    for (const auto& c : r) {
      if (c != 0) {
        return true;
      }
    }
  }
  return false;
}
TSolution Solve(TArray& arr, ll n, ll m, bool inverted) {
  TSolution result;
  for (ll row = 0; row < n; ++row) {
    ll mnRow = arr[row][0];
    for (ll col = 1; col < m; ++col) {
      mnRow = std::min(mnRow, arr[row][col]);
    }
    while (mnRow > 0) {
      result.Rows.push_back(row + 1);
      for (ll col = 0; col < m; ++col) {
        --arr[row][col];
      }
      --mnRow;
    }
  }
  for (ll col = 0; col < m; ++col) {
    ll mnCol = arr[0][col];
    for (ll row = 1; row < n; ++row) {
      mnCol = std::min(mnCol, arr[row][col]);
    }
    while (mnCol > 0) {
      result.Cols.push_back(col + 1);
      for (ll row = 0; row < n; ++row) {
        --arr[row][col];
      }
      --mnCol;
    }
  }
  result.Inverted = inverted;
  result.IsGood = !HasNonZeroes(arr);
  return result;
};
int main() {
  ll n, m;
  std::cin >> n >> m;
  TArray arr(n, std::vector<ll>(m, 0));
  TArray inverted(m, std::vector<ll>(n, 0));
  for (ll row = 0; row < n; ++row) {
    for (ll col = 0; col < m; ++col) {
      std::cin >> arr[row][col];
      inverted[col][row] = arr[row][col];
    }
  }
  if (!HasNonZeroes(arr)) {
    std::cout << "0" << std::endl;
    return 0;
  }
  std::vector<TSolution> slns{
      Solve(arr, n, m, false),
      Solve(inverted, m, n, true),
  };
  const auto printSln = [](const TSolution& sln) {
    ll total = sln.GetTotal();
    if (total == 0 || !sln.IsGood) {
      std::cout << "-1" << std::endl;
    } else {
      std::cout << total << std::endl;
      for (auto r : sln.Inverted ? sln.Cols : sln.Rows) {
        std::cout << "row " << r << std::endl;
      }
      for (auto c : sln.Inverted ? sln.Rows : sln.Cols) {
        std::cout << "col " << c << std::endl;
      }
    }
  };
  printSln(slns[0].IsBetter(slns[1]) ? slns[0] : slns[1]);
}