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;
void readline(char *s) {
  int len = 0;
  char c;
  for (c = ((char)fgetc(stdin)); c != '\n';
       s[len++] = c, c = ((char)fgetc(stdin)))
    ;
}
void readstr(char *s) {
  int len = 0;
  char c;
  for (c = ((char)fgetc(stdin)); c != ' ' && c != '\n';
       s[len++] = c, c = ((char)fgetc(stdin)))
    ;
}
void printstr(char *s) {
  int len = strlen(s);
  for (int(i) = (0); (i) <= (len - 1); i++) fputc((s[i]), stdout);
}
void to_lower(char *s, int L, int R) {
  for (int(i) = (L); (i) <= (R); i++)
    if (((s[i]) >= 'A' && (s[i]) <= 'Z')) s[i] = s[i] - 'A' + 'a';
}
void to_upper(char *s, int L, int R) {
  for (int(i) = (L); (i) <= (R); i++)
    if (((s[i]) >= 'a' && (s[i]) <= 'z')) s[i] = s[i] - 'a' + 'A';
}
long long readll() {
  char c;
  long long ans = 0, x = 1;
  for (c = ((char)fgetc(stdin)); !((c) >= '0' && (c) <= '9') && c != '-';
       c = ((char)fgetc(stdin)))
    ;
  if (c == '-') x = -1, c = ((char)fgetc(stdin));
  for (; ((c) >= '0' && (c) <= '9');
       ans = ans * 10 + c - '0', c = ((char)fgetc(stdin)))
    ;
  return ans * x;
}
void printll(long long x) {
  int a[25], cnt = 0;
  if (x == 0) {
    fputc(('0'), stdout);
    return;
  }
  if (x < 0) fputc(('-'), stdout), x = -x;
  for (; x; a[cnt++] = x - x / 10 * 10, x /= 10)
    ;
  for (int(i) = (cnt - 1); (i) >= (0); i--) fputc((a[i] + '0'), stdout);
}
stringstream ss, ss1;
int n, m, grid[105][105], grid1[105][105];
int main() {
  n = ((int)readll());
  m = ((int)readll());
  for (int(i) = (1); (i) <= (n); i++)
    for (int(j) = (1); (j) <= (m); j++)
      grid[i][j] = ((int)readll()), grid1[i][j] = grid[i][j];
  int ans = 0, ans1 = 0;
  for (int(i) = (1); (i) <= (n); i++) {
    int mn = grid[i][1];
    for (int(j) = (2); (j) <= (m); j++) mn = min(mn, grid[i][j]);
    if (!mn) continue;
    for (int(j) = (1); (j) <= (m); j++) grid[i][j] -= mn;
    for (int(j) = (1); (j) <= (mn); j++) {
      ss << "row"
         << " " << i << endl;
      ans++;
    }
  }
  for (int(i) = (1); (i) <= (m); i++) {
    int mn = grid[1][i];
    for (int(j) = (2); (j) <= (n); j++) mn = min(mn, grid[j][i]);
    if (!mn) continue;
    for (int(j) = (1); (j) <= (n); j++) grid[j][i] -= mn;
    for (int(j) = (1); (j) <= (mn); j++) {
      ss << "col"
         << " " << i << endl;
      ans++;
    }
  }
  bool f1 = true;
  for (int(i) = (1); (i) <= (n); i++)
    for (int(j) = (1); (j) <= (m); j++) {
      if (grid[i][j]) {
        f1 = false;
      }
    }
  for (int(i) = (1); (i) <= (m); i++) {
    int mn = grid1[1][i];
    for (int(j) = (2); (j) <= (n); j++) mn = min(mn, grid1[j][i]);
    if (!mn) continue;
    for (int(j) = (1); (j) <= (n); j++) grid1[j][i] -= mn;
    for (int(j) = (1); (j) <= (mn); j++) {
      ss1 << "col"
          << " " << i << endl;
      ans1++;
    }
  }
  for (int(i) = (1); (i) <= (n); i++) {
    int mn = grid1[i][1];
    for (int(j) = (2); (j) <= (m); j++) mn = min(mn, grid1[i][j]);
    if (!mn) continue;
    for (int(j) = (1); (j) <= (m); j++) grid1[i][j] -= mn;
    for (int(j) = (1); (j) <= (mn); j++) {
      ss1 << "row"
          << " " << i << endl;
      ans1++;
    }
  }
  bool f2 = true;
  for (int(i) = (1); (i) <= (n); i++)
    for (int(j) = (1); (j) <= (m); j++) {
      if (grid[i][j]) {
        f2 = false;
      }
    }
  if (!f1 && !f2) {
    printll((long long)(-1));
  } else {
    if (f1 && !f2) {
      printll((long long)(ans));
      fputc(('\n'), stdout);
      cout << ss.str();
    } else if (f2 && !f1) {
      printll((long long)(ans1));
      fputc(('\n'), stdout);
      cout << ss1.str();
    } else {
      printll((long long)(min(ans, ans1)));
      fputc(('\n'), stdout);
      cout << (ans < ans1 ? ss.str() : ss1.str());
    }
  }
  return 0;
}