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>
long long mpow(long long a, long long n, long long mod) {
  long long ret = 1;
  long long b = a;
  while (n) {
    if (n & 1) ret = (ret * b) % mod;
    b = (b * b) % mod;
    n >>= 1;
  }
  return (long long)ret;
}
inline void read(int &x) {
  x = 0;
  register char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    ;
  for (; c >= '0' && c <= '9'; c = getchar())
    x = (x << 3) + (x << 1) + (c - '0');
}
inline void write(int x) {
  register char buffor[35];
  register int i = 0;
  do {
    buffor[i++] = (x % 10) + '0';
    x /= 10;
  } while (x);
  i--;
  while (i >= 0) putchar(buffor[i--]);
  putchar('\n');
}
inline int in() {
  int x;
  read(x);
  return x;
}
void solve();
int dbg = 1;
using namespace std;
int main() {
  int t = 1;
  for (int i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
int g[3][101][101], arr[101][101], mi[2][101];
string second[3];
int func(int n, int m, int k) {
  int ans = 0;
  for (int i = 0; i < n; i++) {
    ans += mi[k][i];
    for (int j = 0; j < m; j++) {
      arr[i][j] = mi[k][i];
    }
  }
  for (int i = 0; i < m; i++)
    if (arr[0][i] != g[k][0][i]) {
      int diff = g[k][0][i] - arr[0][i];
      for (int j = 0; j < n; j++)
        if ((arr[j][i] + diff) != g[k][j][i]) return -1;
      for (int j = 0; j < n; j++) arr[j][i] += diff;
      ans += diff;
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (arr[i][j] != g[k][i][j]) return -1;
  return ans;
}
void func2(int n, int m, int k) {
  for (int i = 0; i < n; i++) {
    for (int r = 0; r < mi[k][i]; r++)
      cout << second[k] << " " << i + 1 << endl;
    for (int j = 0; j < m; j++) {
      arr[i][j] = mi[k][i];
    }
  }
  for (int i = 0; i < m; i++)
    if (arr[0][i] != g[k][0][i]) {
      int diff = g[k][0][i] - arr[0][i];
      for (int j = 0; j < n; j++) arr[j][i] += diff;
      for (int j = 0; j < diff; j++)
        cout << second[1 - k] << " " << i + 1 << endl;
    }
}
void solve() {
  int n, m;
  cin >> n >> m;
  second[0] = "row";
  second[1] = "col";
  for (int i = 0; i < 101; i++)
    for (int j = 0; j < 2; j++) mi[j][i] = 505;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      cin >> g[0][i][j];
      g[1][j][i] = g[0][i][j];
    }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) {
      mi[0][i] = min(mi[0][i], g[0][i][j]);
      mi[1][j] = min(mi[1][j], g[0][i][j]);
    }
  int k, k2;
  k = func(n, m, 0);
  k2 = func(m, n, 1);
  if (k >= 0) {
    cout << min(k, k2) << endl;
    if (k < k2)
      func2(n, m, 0);
    else
      func2(m, n, 1);
    return;
  }
  cout << -1 << endl;
  ;
}