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;
signed main() {
  ios::sync_with_stdio(false);
  long long t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    long long a[n][m];
    vector<long long> r(n + 1, 0), c(m + 1, 0), m_r(n + 1, INT_MAX),
        m_c(m + 1, INT_MAX);
    long long row = INT_MAX, col = INT_MAX;
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < m; j++) cin >> a[i][j];
    for (long long i = 0; i < n; i++)
      for (long long j = 0; j < m; j++) m_r[i + 1] = min(m_r[i + 1], a[i][j]);
    for (long long i = 0; i < m; i++)
      for (long long j = 0; j < n; j++) m_c[i + 1] = min(m_c[i + 1], a[j][i]);
    vector<vector<long long> > v(n + 1, vector<long long>(m + 1, 0));
    if (n <= m) {
      for (long long i = 0; i < n; i++) {
        if (m_r[i + 1] >= 1) {
          for (long long j = 0; j < m; j++) v[i][j] += m_r[i + 1];
          r[i + 1] = m_r[i + 1];
        }
        row = min(row, m_r[i + 1]);
      }
      for (long long i = 0; i < m; i++) {
        if (row < m_c[i + 1]) {
          for (long long j = 0; j < n; j++) v[j][i] += m_c[i + 1] - row;
          c[i + 1] = m_c[i + 1] - row;
        }
      }
    } else {
      for (long long i = 0; i < m; i++) {
        if (m_c[i + 1] >= 1) {
          for (long long j = 0; j < n; j++) v[j][i] += m_c[i + 1];
          c[i + 1] = m_c[i + 1];
        }
        col = min(col, m_c[i + 1]);
      }
      for (long long i = 0; i < n; i++) {
        if (col < m_r[i + 1]) {
          for (long long j = 0; j < m; j++) v[i][j] += m_r[i + 1] - col;
          r[i + 1] = m_r[i + 1] - col;
        }
      }
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        if (v[i][j] != a[i][j]) {
          cout << -1;
          return 0;
        }
      }
    }
    long long ans = 0;
    for (long long i = 0; i < n; i++)
      if (r[i + 1] != 0) ans += r[i + 1];
    for (long long i = 0; i < m; i++)
      if (c[i + 1] != 0) ans += c[i + 1];
    cout << ans << endl;
    for (long long i = 0; i < n; i++) {
      if (r[i + 1] != 0) {
        for (long long j = 0; j < r[i + 1]; j++) {
          cout << "row " << i + 1 << endl;
        }
      }
    }
    for (long long i = 0; i < m; i++) {
      if (c[i + 1] != 0) {
        for (long long j = 0; j < c[i + 1]; j++) {
          cout << "col " << i + 1 << endl;
        }
      }
    }
  }
}