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;
multiset<pair<int, int> > r[107], c[107];
vector<pair<int, int> > ans;
void imp() { printf("-1"); }
void solve() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int(i) = 0; (i) < (n); ++(i))
    for (int(j) = 0; (j) < (m); ++(j)) {
      int a;
      scanf("%d", &a);
      r[i].insert({a, j});
      c[j].insert({a, i});
    }
  for (int iter = 0; iter <= 50005; ++iter) {
    bool any = false;
    for (int(i) = 0; (i) < (n); ++(i))
      if (r[i].begin()->first < r[i].rbegin()->first) {
        any = true;
        int j = r[i].rbegin()->second;
        ans.push_back({1, j + 1});
        multiset<pair<int, int> > nw;
        for (auto it = c[j].begin(); it != c[j].end(); ++it) {
          if (it->first == 0) return imp();
          nw.insert({(it->first) - 1, it->second});
          int pos = it->second;
          r[pos].erase(r[pos].find({it->first, j}));
          r[pos].insert({(it->first) - 1, j});
        }
        c[j] = nw;
        break;
      }
    if (!any) break;
  }
  int mn = 700;
  for (int(i) = 0; (i) < (n); ++(i)) mn = min(mn, r[i].begin()->first);
  for (int(i) = 0; (i) < (n); ++(i)) {
    int to = r[i].begin()->first - mn;
    while (to--) ans.push_back({0, i + 1});
  }
  if (n <= m)
    for (int(i) = 0; (i) < (n); ++(i))
      for (int(j) = 0; (j) < (mn); ++(j)) ans.push_back({0, i + 1});
  else
    for (int(j) = 0; (j) < (m); ++(j))
      for (int(i) = 0; (i) < (mn); ++(i)) ans.push_back({1, j + 1});
  printf("%d\n", (int)(ans).size());
  for (pair<int, int>& i : ans) {
    if (i.first == 0)
      printf("row ");
    else
      printf("col ");
    printf("%d\n", i.second);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}
