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:

<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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline void R(int &v) {
  static char ch;
  v = 0;
  bool p = 0;
  do {
    ch = getchar();
    if (ch == '-') p = 1;
  } while (!isdigit(ch));
  while (isdigit(ch)) {
    v = (v + (v << 2) << 1) + (ch ^ '0');
    ch = getchar();
  }
  if (p) v = -v;
}
int ret;
struct daan {
  int kind, num;
} ans[1000005];
int jishu;
int ma[105][105];
int n, m;
bool pd;
int inf;
inline void caozuo(int i, int j) {
  int minheng = inf, minshu = inf;
  for (int k = 1; k <= m; ++k) {
    if (k != j) minheng = min(minheng, ma[i][k]);
  }
  for (int k = 1; k <= n; ++k) {
    if (k != i) minshu = min(minshu, ma[k][j]);
  }
  if (ma[i][j] > 0 && minheng == 0 && minshu == 0) {
    pd = 0;
    return;
  }
  while (minheng > 0 && ma[i][j] > 0) {
    for (int k = 1; k <= m; ++k) {
      ma[i][k]--;
    }
    minheng--;
    ans[++jishu].kind = 1, ans[jishu].num = i;
  }
  while (minshu > 0 && ma[i][j] > 0) {
    for (int k = 1; k <= n; ++k) {
      ma[k][j]--;
    }
    minshu--;
    ans[++jishu].kind = 2, ans[jishu].num = j;
  }
}
inline void caozuo1(int i, int j) {
  int minheng = inf, minshu = inf;
  for (int k = 1; k <= m; ++k) {
    if (k != j) minheng = min(minheng, ma[i][k]);
  }
  for (int k = 1; k <= n; ++k) {
    if (k != i) minshu = min(minshu, ma[k][j]);
  }
  if (ma[i][j] > 0 && minheng == 0 && minshu == 0) {
    pd = 0;
    return;
  }
  while (minshu > 0 && ma[i][j] > 0) {
    for (int k = 1; k <= n; ++k) {
      ma[k][j]--;
    }
    minshu--;
    ans[++jishu].kind = 2, ans[jishu].num = j;
  }
  while (minheng > 0 && ma[i][j] > 0) {
    for (int k = 1; k <= m; ++k) {
      ma[i][k]--;
    }
    minheng--;
    ans[++jishu].kind = 1, ans[jishu].num = i;
  }
}
int main() {
  R(n), R(m);
  pd = 1;
  memset(ma, 127, sizeof(ma));
  inf = ma[0][0];
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      R(ma[i][j]);
    }
  }
  if (m >= n) {
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        caozuo(i, j);
        if (!pd) {
          puts("-1");
          return 0;
        }
      }
    }
  } else {
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        caozuo1(i, j);
        if (!pd) {
          puts("-1");
          return 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (ma[i][j]) {
        puts("-1");
        return 0;
      }
    }
  }
  printf("%d\n", jishu);
  for (int i = 1; i <= jishu; ++i) {
    if (ans[i].kind == 1) {
      printf("row %d\n", ans[i].num);
    } else
      printf("col %d\n", ans[i].num);
  }
  return 0;
}