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:

<image>

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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  std::ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  int A[n][m];
  int B[n][m];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> A[i][j];
      B[i][j] = 0;
    }
  }
  if (m >= n) {
    vector<string> ans_1;
    vector<int> ans_12;
    for (int i = 0; i < n; i++) {
      int nim = 1e6, in1 = -1, in2 = -1;
      for (int j = 0; j < m; j++) {
        if (A[i][j] < nim) {
          nim = A[i][j];
          in1 = i;
          in2 = j;
        }
      }
      nim = A[in1][in2] - B[in1][in2];
      if (nim > 0 && B[in1][in2] < A[in1][in2]) {
        char q = i + 1;
        int t = 0;
        while (t < nim) {
          ans_1.push_back("row ");
          ans_12.push_back(q);
          t++;
        }
        for (int j = 0; j < m; j++) {
          B[i][j] += nim;
        }
      }
    }
    for (int i = 0; i < m; i++) {
      int nim = 1e6, in1 = -1, in2 = -1;
      for (int j = 0; j < n; j++) {
        if (A[j][i] < nim) {
          nim = A[j][i];
          in1 = j;
          in2 = i;
        }
      }
      nim = A[in1][in2] - B[in1][in2];
      if (nim > 0 && B[in1][in2] < A[in1][in2]) {
        char q = i + 1;
        int t = 0;
        while (t < nim) {
          ans_1.push_back("col ");
          ans_12.push_back(q);
          t++;
        }
        for (int j = 0; j < n; j++) {
          B[j][i] += nim;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (A[i][j] != B[i][j]) {
          cout << -1;
          return 0;
        }
      }
    }
    cout << ans_1.size() << "\n";
    for (int i = 0; i < ans_1.size(); i++) {
      cout << ans_1[i] << ans_12[i];
      if (i != ans_1.size() - 1) {
        cout << "\n";
      }
    }
  } else {
    vector<string> ans_1;
    vector<int> ans_12;
    for (int i = 0; i < m; i++) {
      int nim = 1e6, in1 = -1, in2 = -1;
      for (int j = 0; j < n; j++) {
        if (A[j][i] < nim) {
          nim = A[j][i];
          in1 = j;
          in2 = i;
        }
      }
      nim = A[in1][in2] - B[in1][in2];
      if (nim > 0 && B[in1][in2] < A[in1][in2]) {
        char q = i + 1;
        int t = 0;
        while (t < nim) {
          ans_1.push_back("col ");
          ans_12.push_back(q);
          t++;
        }
        for (int j = 0; j < n; j++) {
          B[j][i] += nim;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      int nim = 1e6, in1 = -1, in2 = -1;
      for (int j = 0; j < m; j++) {
        if (A[i][j] < nim) {
          nim = A[i][j];
          in1 = i;
          in2 = j;
        }
      }
      nim = A[in1][in2] - B[in1][in2];
      if (nim > 0 && B[in1][in2] < A[in1][in2]) {
        char q = i + 1;
        int t = 0;
        while (t < nim) {
          ans_1.push_back("row ");
          ans_12.push_back(q);
          t++;
        }
        for (int j = 0; j < m; j++) {
          B[i][j] += nim;
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        if (A[i][j] != B[i][j]) {
          cout << -1;
          return 0;
        }
      }
    }
    cout << ans_1.size() << "\n";
    for (int i = 0; i < ans_1.size(); i++) {
      cout << ans_1[i] << ans_12[i];
      if (i != ans_1.size() - 1) {
        cout << "\n";
      }
    }
  }
  return 0;
}