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;
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    T t = bigmod(p, e / 2, M);
    return (t * t) % M;
  }
  return (bigmod(p, e - 1, M) * p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T multimod(T a, T b, T m) {
  T ans = 0ll;
  a %= m, b %= m;
  while (b) {
    if (b & 1ll) ans = m - ans > a ? (ans + a) : (ans + a - m);
    b >>= 1ll;
    a = (m - a) > a ? a + a : a + a - m;
  }
  return (T)ans;
}
void sc(int &a) { scanf("%d", &a); }
void sc(long long &a) { scanf("%lld", &a); }
void sc(double &a) { scanf("%lf", &a); }
void sc(int &a, int &b) { scanf("%d %d", &a, &b); }
void sc(long long &a, long long &b) { scanf("%lld %lld", &a, &b); }
void sc(int &a, int &b, int &c) { scanf("%d %d %d", &a, &b, &c); }
void sc(int &a, int &b, long long &c) { scanf("%d %d %lld", &a, &b, &c); }
void sc(long long &a, long long &b, long long &c) {
  scanf("%lld %lld %lld", &a, &b, &c);
}
void sc(string &str) { cin >> str; }
void sc(char *(str)) { scanf(" %s", str); }
void sc(char &c) { scanf(" %c", &c); }
int arr[110][110];
int main() {
  int i, j, k;
  int t, cases = 1;
  string str;
  int q, u, n, v, m;
  sc(n, m);
  vector<int> ans;
  if (n < m) {
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) sc(arr[i][j]);
    }
    for (i = 0; i < n; i++) {
      while (1) {
        for (j = 0; j < m; j++) {
          if (arr[i][j] <= 0) break;
        }
        if (j == m) {
          ans.push_back(-1);
          ans.push_back(i);
          for (j = 0; j < m; j++) arr[i][j]--;
        } else
          break;
      }
    }
    for (i = 0; i < m; i++) {
      while (1) {
        for (j = 0; j < n; j++) {
          if (arr[j][i] <= 0) break;
        }
        if (j == n) {
          ans.push_back(-2);
          ans.push_back(i);
          for (j = 0; j < n; j++) arr[j][i]--;
        } else
          break;
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (arr[i][j] != 0) {
          cout << -1 << endl;
          return 0;
        }
      }
    }
    int sz = ans.size();
    cout << sz / 2 << endl;
    for (i = 0; i < sz; i += 2) {
      if (ans[i] == -1)
        printf("row ");
      else
        printf("col ");
      printf("%d\n", ans[i + 1] + 1);
    }
  } else {
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) sc(arr[i][j]);
    }
    for (i = 0; i < m; i++) {
      while (1) {
        for (j = 0; j < n; j++) {
          if (arr[j][i] <= 0) break;
        }
        if (j == n) {
          ans.push_back(-2);
          ans.push_back(i);
          for (j = 0; j < n; j++) arr[j][i]--;
        } else
          break;
      }
    }
    for (i = 0; i < n; i++) {
      while (1) {
        for (j = 0; j < m; j++) {
          if (arr[i][j] <= 0) break;
        }
        if (j == m) {
          ans.push_back(-1);
          ans.push_back(i);
          for (j = 0; j < m; j++) arr[i][j]--;
        } else
          break;
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        if (arr[i][j] != 0) {
          cout << -1 << endl;
          return 0;
        }
      }
    }
    int sz = ans.size();
    cout << sz / 2 << endl;
    for (i = 0; i < sz; i += 2) {
      if (ans[i] == -1)
        printf("row ");
      else
        printf("col ");
      printf("%d\n", ans[i + 1] + 1);
    }
  }
  return 0;
}