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.
#include <bits/stdc++.h>
using namespace std;
void SieveOfErat() {
  bool prime[1000001];
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (long long i = p * 2; i <= 1000000; i += p) prime[i] = false;
    }
  }
}
long long gcd(long long x, long long y) {
  if (x == 0) {
    return y;
  } else
    return gcd(x % y, y);
}
long long n, m;
vector<vector<long long> > v;
vector<long long> res;
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  v.assign(n, vector<long long>(m, 0));
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < m; ++j) cin >> v[i][j];
  }
  if (n < m) {
    for (long long i = 0; i < n; ++i) {
      long long mi = 600;
      for (long long j = 0; j < m; ++j) mi = min(mi, v[i][j]);
      if (mi == 0) continue;
      for (long long j = 0; j < mi; ++j) res.push_back(i + 1);
      for (long long j = 0; j < m; ++j) v[i][j] -= mi;
    }
    for (long long j = 0; j < m; ++j) {
      long long mi = 600;
      for (long long i = 0; i < n; ++i) mi = min(mi, v[i][j]);
      if (mi == 0) continue;
      for (long long i = 0; i < mi; ++i) res.push_back(-(j + 1));
      for (long long i = 0; i < n; ++i) v[i][j] -= mi;
    }
  } else {
    for (long long j = 0; j < m; ++j) {
      long long mi = 600;
      for (long long i = 0; i < n; ++i) mi = min(mi, v[i][j]);
      if (mi == 0) continue;
      for (long long i = 0; i < mi; ++i) res.push_back(-(j + 1));
      for (long long i = 0; i < n; ++i) v[i][j] -= mi;
    }
    for (long long i = 0; i < n; ++i) {
      long long mi = 600;
      for (long long j = 0; j < m; ++j) mi = min(mi, v[i][j]);
      if (mi == 0) continue;
      for (long long j = 0; j < mi; ++j) res.push_back(i + 1);
      for (long long j = 0; j < m; ++j) v[i][j] -= mi;
    }
  }
  long long sum = 0;
  for (long long i = 0; i < n; ++i)
    for (long long j = 0; j < m; ++j) sum += v[i][j];
  if (sum > 0) {
    cout << -1 << endl;
    return 0;
  }
  cout << res.size() << "\n";
  for (long long i = 0; i < res.size(); ++i) {
    if (res[i] > 0)
      cout << "row " << res[i] << "\n";
    else
      cout << "col " << -res[i] << "\n";
  }
  return 0;
}
