Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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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:

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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;
const int nmax = 100;
const int inf = 1 << 30;
int v[nmax + 1][nmax + 1], aux[nmax + 1][nmax + 1];
vector<pair<const char*, int> > sol[2];
void linii(int n, int m, int ind) {
  for (int i = 1; i <= n; ++i) {
    int mn = inf;
    for (int j = 1; j <= m; ++j) mn = min(mn, v[i][j]);
    for (int j = 1; j <= m; ++j) v[i][j] -= mn;
    for (int j = 1; j <= mn; ++j) sol[ind].push_back(make_pair("row", i));
  }
}
void coloane(int n, int m, int ind) {
  for (int j = 1; j <= m; ++j) {
    int mn = inf;
    for (int i = 1; i <= n; ++i) mn = min(mn, v[i][j]);
    for (int i = 1; i <= n; ++i) v[i][j] -= mn;
    for (int i = 1; i <= mn; ++i) sol[ind].push_back(make_pair("col", j));
  }
}
bool solve(int n, int m, int ind) {
  memcpy(v, aux, sizeof(aux));
  if (ind == 0) {
    linii(n, m, ind);
    coloane(n, m, ind);
  } else {
    coloane(n, m, ind);
    linii(n, m, ind);
  }
  bool ok = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (v[i][j]) ok = 0;
    }
  }
  return ok;
}
int main() {
  int n, m;
  fscanf(stdin, "%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      fscanf(stdin, "%d", &aux[i][j]);
    }
  }
  bool o1 = solve(n, m, 0);
  bool o2 = solve(n, m, 1);
  int ans = 0;
  if (o1 == 0 || (o2 != 0 && sol[0].size() > sol[1].size())) ans = 1;
  if (o1 == 0 && o2 == 0) {
    fprintf(stdout, "-1");
  } else {
    fprintf(stdout, "%d\n", (int)sol[ans].size());
    for (auto i : sol[ans]) {
      fprintf(stdout, "%s %d\n", i.first, i.second);
    }
  }
  return 0;
}