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>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
int a[110][110];
int n, m, g[111][111], row[111], col[111], mnrow[111], mncol[111];
vector<pair<int, int> > ans;
bool checkc(int x, int y) {
  int cnt = a[x][y];
  int flag = 1;
  for (int i = 1; i <= m; i++) {
    if (a[x][i] < cnt) flag = 0;
  }
  if (flag) {
    for (int i = 1; i <= cnt; i++) ans.push_back(pair<int, int>(0, x));
    for (int i = 1; i <= m; i++) a[x][i] -= cnt;
  }
  return flag;
}
bool checkl(int x, int y) {
  int cnt = a[x][y];
  int flag = 1;
  for (int i = 1; i <= n; i++) {
    if (a[i][y] < cnt) flag = 0;
  }
  if (flag) {
    for (int i = 1; i <= cnt; i++) ans.push_back(pair<int, int>(1, y));
    for (int i = 1; i <= n; i++) a[i][y] -= cnt;
  }
  return flag;
}
char aa[2][10] = {"row", "col"};
bool cg = 0;
vector<pair<int, int> > ret;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= 111; ++i) mnrow[i] = mncol[i] = 555;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      scanf("%d", &g[i][j]);
      mnrow[i] = min(mnrow[i], g[i][j]);
      mncol[j] = min(mncol[j], g[i][j]);
    }
  }
  int ans = 0;
  if (n >= m) {
    for (int j = 1; j <= m; ++j) {
      int t = mncol[j];
      if (t <= 0) continue;
      for (int i = 1; i <= n; ++i) {
        g[i][j] -= t;
      }
      for (int k = 0; k < t; ++k) {
        ret.push_back({1, j});
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        mnrow[i] = min(mnrow[i], g[i][j]);
      }
    }
    for (int i = 1; i <= n; ++i) {
      int t = mnrow[i];
      if (t <= 0) continue;
      for (int j = 1; j <= m; ++j) {
        g[i][j] -= t;
      }
      for (int k = 0; k < t; ++k) {
        ret.push_back({0, i});
      }
    }
  } else {
    for (int i = 1; i <= n; ++i) {
      int t = mnrow[i];
      if (t <= 0) continue;
      for (int j = 1; j <= m; ++j) {
        g[i][j] -= t;
      }
      for (int k = 0; k < t; ++k) {
        ret.push_back({0, i});
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= m; ++j) {
        mncol[j] = min(mncol[j], g[i][j]);
      }
    }
    for (int j = 1; j <= m; ++j) {
      int t = mncol[j];
      if (t <= 0) continue;
      for (int i = 1; i <= n; ++i) {
        g[i][j] -= t;
      }
      for (int k = 0; k < t; ++k) {
        ret.push_back({1, j});
      }
    }
  }
  bool ok = true;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= m; ++j) {
      if (g[i][j] != 0) ok = false;
    }
  }
  if (!ok)
    puts("-1");
  else {
    cout << ret.size() << endl;
    for (auto it : ret) {
      printf("%s %d\n", aa[it.first], it.second);
    }
  }
  return 0;
}