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;
long long int pwr(long long int base, long long int p, long long int mod) {
  long long int ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
vector<pair<int, int>> v;
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  int a[n][m];
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      scanf("%d", &a[i][j]);
    }
  }
  int minn = INT_MAX;
  for (int i = (0); i < (m); ++i) {
    minn = min(a[0][i], minn);
  }
  for (int i = (0); i < (m); ++i) {
    while (a[0][i] != minn) {
      for (int j = (0); j < (n); ++j) {
        a[j][i]--;
      }
      v.push_back(make_pair(1, i + 1));
    }
  }
  minn = INT_MAX;
  for (int i = (0); i < (n); ++i) {
    minn = min(a[i][0], minn);
  }
  for (int i = (0); i < (n); ++i) {
    while (a[i][0] != minn) {
      for (int j = (0); j < (m); ++j) {
        a[i][j]--;
      }
      v.push_back(make_pair(0, i + 1));
    }
  }
  int x = a[0][0];
  if (n >= m) {
    for (int i = (0); i < (m); ++i) {
      for (int j = (0); j < (x); ++j) {
        v.push_back(make_pair(1, i + 1));
      }
    }
  } else {
    for (int i = (0); i < (n); ++i) {
      for (int j = (0); j < (x); ++j) {
        v.push_back(make_pair(0, i + 1));
      }
    }
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      a[i][j] -= x;
    }
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      if (a[i][j]) {
        cout << -1;
        return 0;
      }
    }
  }
  int l = v.size();
  printf("%d\n", l);
  for (int i = (0); i < (l); ++i) {
    if (v[i].first)
      cout << "col ";
    else
      cout << "row ";
    cout << v[i].second << endl;
  }
  return 0;
}
