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;
int A[200][200], B[200][200], C[200][200];
int tmp[200][200], n, m;
vector<string> res;
bool check() {
  bool ret = 1;
  for (int i = 0; i < n; i++)
    for (int ii = 0; ii < m; ii++)
      if (B[i][ii] != C[i][ii]) ret = 0;
  return ret;
}
void solveCol() {
  for (int i = 0; i < m; i++) {
    int mn = 10000;
    for (int ii = 0; ii < n; ii++) mn = min(mn, A[ii][i]);
    for (int ii = 0; ii < mn; ii++) res.push_back("col " + to_string(i + 1));
    for (int ii = 0; ii < n; ii++) A[ii][i] -= mn, B[ii][i] += mn;
  }
}
void solveRow() {
  for (int i = 0; i < n; i++) {
    int mn = 10000;
    for (int ii = 0; ii < m; ii++) mn = min(mn, A[i][ii]);
    for (int ii = 0; ii < mn; ii++) res.push_back("row " + to_string(i + 1));
    for (int ii = 0; ii < m; ii++) A[i][ii] -= mn, B[i][ii] += mn;
  }
}
bool solve() {
  for (int i = 0; i < n; i++)
    for (int ii = 0; ii < m; ii++) tmp[i][ii] = A[i][ii];
  solveRow();
  solveCol();
  vector<string> r1 = res;
  int mn1 = 1e9;
  if (check()) mn1 = r1.size();
  bool ret = 1;
  memset(B, 0, sizeof B);
  for (int i = 0; i < n; i++)
    for (int ii = 0; ii < m; ii++) A[i][ii] = tmp[i][ii];
  res.clear();
  solveCol();
  solveRow();
  if (check()) {
    if (res.size() > mn1) res = r1;
    return 1;
  }
  if (mn1 != 1e9) {
    res = r1;
    return 1;
  }
  return 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++)
    for (int ii = 0; ii < m; ii++) scanf("%d", &A[i][ii]), C[i][ii] = A[i][ii];
  bool ret = solve();
  if (ret) {
    printf("%d\n", res.size());
    for (int i = 0; i < res.size(); i++) printf("%s\n", res[i].c_str());
  } else
    puts("-1");
}