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;
const int MAX_N = 106;
int g[MAX_N][MAX_N];
int h[MAX_N][MAX_N];
bool not_empty_row[MAX_N];
bool not_empty_col[MAX_N];
bool used_row[MAX_N];
bool used_col[MAX_N];
void coppy(int n, int m) {
  for (int i = 1; n >= i; i++) {
    for (int j = 1; m >= j; j++) {
      h[i][j] = g[i][j];
    }
  }
  memset(used_row, 0, sizeof(used_row));
  memset(used_col, 0, sizeof(used_col));
}
bool okayy(int n, int m) {
  bool ret = true;
  for (int i = 1; n >= i; i++) {
    for (int j = 1; m >= j; j++) {
      ret &= (h[i][j] == 0);
    }
  }
  return ret;
}
bool deleted(int type, int x, int val, int n, int m) {
  if (type == 1) {
    int i = x;
    for (int j = 1; m >= j; j++) {
      h[i][j] -= val;
      if (h[i][j] < 0) return false;
    }
    return true;
  }
  if (type == 2) {
    int j = x;
    for (int i = 1; n >= i; i++) {
      h[i][j] -= val;
      if (h[i][j] < 0) return false;
    }
    return true;
  }
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  bool not_empty = false;
  for (int i = 1; n >= i; i++) {
    for (int j = 1; m >= j; j++) {
      scanf("%d", &g[i][j]);
      if (g[i][j]) {
        not_empty_row[i] = 1;
        not_empty_col[j] = 1;
        not_empty = true;
      }
    }
  }
  if (!not_empty) {
    puts("0");
    return 0;
  }
  int rrow = -1;
  for (int i = 1; n >= i; i++) {
    if (not_empty_row[i]) {
      rrow = i;
      break;
    }
  }
  if (rrow == -1) cout << 0 / 0 << endl;
  vector<pair<int, pair<int, int> > > ans;
  vector<pair<int, pair<int, int> > > real_ans;
  int meruru = 123456789;
  int xx = 0;
  for (int cnt = 0; g[rrow][1] >= cnt; cnt++) {
    ans.clear();
    coppy(n, m);
    bool ret = deleted(1, rrow, cnt, n, m);
    xx = 0;
    xx += cnt;
    ans.push_back(make_pair(1, make_pair(rrow, cnt)));
    for (int j = 1; m >= j; j++) {
      xx += h[rrow][j];
      ans.push_back(make_pair(2, make_pair(j, h[rrow][j])));
      ret &= deleted(2, j, h[rrow][j], n, m);
    }
    for (int i = 1; n >= i; i++) {
      xx += h[i][1];
      ans.push_back(make_pair(1, make_pair(i, h[i][1])));
      ret &= deleted(1, i, h[i][1], n, m);
    }
    if (ret && okayy(n, m)) {
      if (xx < meruru) {
        real_ans = ans;
        meruru = xx;
      }
    } else
      ans.clear();
  }
  ans = real_ans;
  if (ans.size() == 0)
    puts("-1");
  else {
    int cnt = 0;
    for (pair<int, pair<int, int> > i : ans) {
      cnt += i.second.second;
    }
    if (cnt != meruru) cout << 0 / 0 << endl;
    printf("%d\n", cnt);
    for (pair<int, pair<int, int> > i : ans) {
      while (i.second.second--) {
        if (i.first == 1)
          printf("row ");
        else
          printf("col ");
        printf("%d\n", i.second.first);
      }
    }
  }
}