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;
inline void smax(int &x, int y) { x = max(x, y); }
inline void smin(int &x, int y) { x = min(x, y); }
inline int in() {
  int32_t x;
  scanf("%d", &x);
  return x;
}
inline long long lin() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
inline void out(int x) { printf("%d ", x); }
inline void lout(long long x) { printf("%I64d ", x); }
const int MOD = 1e9 + 7;
const int INF = 1e9;
const long long LINF = 1e18;
const int N = 1e3 + 10;
const int maxn = 1e4 + 10;
const int MAXN = 1e5 + 10;
const int MN = 1e6 + 10;
int a[110][110], mn[2][110], n, m, mx[110];
vector<pair<string, int> > vec;
void solve() {
  fill_n(mn[0], 2 * 110, INF);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) smin(mn[0][i], a[i][j]);
  for (int j = 1; j <= m; j++)
    for (int i = 1; i <= n; i++) smin(mn[1][j], a[i][j]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++) cin >> a[i][j];
  solve();
  if (n < m) {
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < mn[0][i]; j++) vec.push_back({"row", i});
      for (int j = 1; j <= m; j++) a[i][j] -= mn[0][i];
    }
    solve();
    for (int i = 1; i <= m; i++) {
      for (int j = 0; j < mn[1][i]; j++) vec.push_back({"col", i});
      for (int j = 1; j <= n; j++) a[j][i] -= mn[1][i];
    }
  } else {
    for (int i = 1; i <= m; i++) {
      for (int j = 0; j < mn[1][i]; j++) vec.push_back({"col", i});
      for (int j = 1; j <= n; j++) a[j][i] -= mn[1][i];
    }
    solve();
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < mn[0][i]; j++) vec.push_back({"row", i});
      for (int j = 1; j <= m; j++) a[i][j] -= mn[0][i];
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= m; j++)
      if (a[i][j] != 0) return cout << -1, 0;
  cout << vec.size() << "\n";
  for (auto x : vec) cout << x.first << ' ' << x.second << "\n";
}