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;
int tar[111][111];
int arr[111][111];
int need[111][111];
vector<pair<string, int> > vec1;
vector<pair<string, int> > vec2;
int row, col;
bool done() {
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) {
      if (arr[i][j] != tar[i][j]) return false;
    }
  }
  return true;
}
void seeArr() {
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) {
      cout << arr[i][j] << " ";
    }
    cout << "\n";
  }
  cout << "\n";
}
void makeNeed() {
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) need[i][j] = tar[i][j] - arr[i][j];
  }
}
void fillRow() {
  memset(arr, 0, sizeof(arr));
  for (int i = 0; i < row; i++) {
    if (done()) return;
    int mn = 505;
    for (int j = 0; j < col; j++) mn = min(mn, tar[i][j]);
    for (int j = 0; j < col; j++) arr[i][j] += mn;
    while (mn--) vec1.push_back(make_pair("row", i + 1));
  }
  makeNeed();
  for (int i = 0; i < col; i++) {
    if (done()) return;
    int mn = 505;
    for (int j = 0; j < row; j++) mn = min(mn, need[j][i]);
    for (int j = 0; j < row; j++) arr[j][i] += mn;
    while (mn--) vec1.push_back(make_pair("col", i + 1));
  }
}
void fillCol() {
  memset(arr, 0, sizeof(arr));
  for (int i = 0; i < col; i++) {
    if (done()) return;
    int mn = 505;
    for (int j = 0; j < row; j++) mn = min(mn, tar[j][i]);
    for (int j = 0; j < row; j++) arr[j][i] += mn;
    while (mn--) vec2.push_back(make_pair("col", i + 1));
  }
  makeNeed();
  for (int i = 0; i < row; i++) {
    if (done()) return;
    int mn = 505;
    for (int j = 0; j < col; j++) mn = min(mn, need[i][j]);
    for (int j = 0; j < col; j++) arr[i][j] += mn;
    while (mn--) vec2.push_back(make_pair("row", i + 1));
  }
}
void printVec1() {
  cout << vec1.size() << "\n";
  for (int i = 0; i < vec1.size(); i++)
    cout << vec1[i].first << " " << vec1[i].second << "\n";
}
void printVec2() {
  cout << vec2.size() << "\n";
  for (int i = 0; i < vec2.size(); i++)
    cout << vec2[i].first << " " << vec2[i].second << "\n";
}
int main() {
  cin >> row;
  cin >> col;
  memset(arr, 0, sizeof(arr));
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) cin >> tar[i][j];
  }
  fillRow();
  bool rowErJonno = done();
  fillCol();
  bool colErJonno = done();
  if (rowErJonno && colErJonno) {
    if (vec1.size() <= vec2.size())
      printVec1();
    else
      printVec2();
  } else if (rowErJonno)
    printVec1();
  else if (colErJonno)
    printVec2();
  else
    cout << -1 << "\n";
  return 0;
}