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.
#include <bits/stdc++.h>
using namespace std;
long long GCD(long long n1, long long n2) {
  if (n2 != 0)
    return GCD(n2, n1 % n2);
  else
    return n1;
}
long long LCM(long long n1, long long n2) {
  long long m;
  m = n1 * n2 / GCD(n1, n2);
  return m;
}
unsigned long long choose(unsigned long long n, unsigned long long k) {
  if (k > n) {
    return 0;
  }
  unsigned long long r = 1;
  for (unsigned long long d = 1; d <= k; ++d) {
    r *= n--;
    r /= d;
  }
  return r;
}
bool isprime(int x) {
  if (x == 1) return 0;
  for (int i = 2; i * i < x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long mod = 1000000007;
long long pow_(long long base, long long exponent, int modulus) {
  long long result = 1;
  while (exponent > 0) {
    if (exponent % 2 == 1) result = (result * base) % mod;
    exponent = exponent >> 1;
    base = (base * base) % mod;
  }
  return result;
}
bool pal(string s) {
  string a = "";
  for (int i = s.size() - 1; i >= 0; i--) a += s[i];
  if (a == s)
    return 1;
  else
    return 0;
}
int arr[105][105], brr[105][105];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<pair<int, int> > vec, tec, rec, cec;
  int x, n, m;
  bool da1 = 1, da2 = 1;
  int sum1 = 0, sum2 = 0;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
      brr[i][j] = arr[i][j];
    }
  }
  int mn = 1e9;
  for (int i = 0; i < n; i++) {
    mn = 1e9;
    for (int j = 0; j < m; j++) {
      mn = min(arr[i][j], mn);
    }
    for (int j = 0; j < m; j++) {
      arr[i][j] -= mn;
    }
    if (mn != 0) {
      sum1 += mn;
      vec.push_back({mn, i + 1});
    }
  }
  for (int i = 0; i < m; i++) {
    mn = arr[0][i];
    for (int j = 1; j < n; j++) {
      if (arr[j][i] != arr[j - 1][i]) {
        da1 = 0;
      }
      mn = min(arr[j][i], mn);
    }
    if (!da1) break;
    if (mn != 0) {
      sum1 += mn;
      tec.push_back({mn, i + 1});
    }
  }
  mn = 1e9;
  for (int i = 0; i < m; i++) {
    mn = 1e9;
    for (int j = 0; j < n; j++) {
      mn = min(brr[j][i], mn);
    }
    for (int j = 0; j < n; j++) {
      brr[j][i] -= mn;
    }
    if (mn != 0) {
      sum2 += mn;
      rec.push_back({mn, i + 1});
    }
  }
  for (int i = 0; i < n; i++) {
    mn = brr[i][0];
    for (int j = 1; j < m; j++) {
      if (brr[i][j] != brr[i][j - 1]) {
        da2 = 0;
      }
      mn = min(brr[i][j], mn);
    }
    if (!da2) break;
    if (mn != 0) {
      sum2 += mn;
      cec.push_back({mn, i + 1});
    }
  }
  int a;
  if (da1 && !da2) {
    cout << sum1 << endl;
    for (int i = 0; i < vec.size(); i++) {
      a = vec[i].first;
      while (a--) {
        cout << "row " << vec[i].second << endl;
      }
    }
    for (int i = 0; i < tec.size(); i++) {
      a = tec[i].first;
      while (a--) {
        cout << "col " << tec[i].second << endl;
      }
    }
  } else if (!da1 && da2) {
    cout << sum2 << endl;
    for (int i = 0; i < rec.size(); i++) {
      a = rec[i].first;
      while (a--) {
        cout << "col " << rec[i].second << endl;
      }
    }
    for (int i = 0; i < cec.size(); i++) {
      a = cec[i].first;
      while (a--) {
        cout << "row " << cec[i].second << endl;
      }
    }
  } else if (da1 && da2) {
    if (sum1 <= sum2) {
      cout << sum1 << endl;
      for (int i = 0; i < vec.size(); i++) {
        a = vec[i].first;
        while (a--) {
          cout << "row " << vec[i].second << endl;
        }
      }
      for (int i = 0; i < tec.size(); i++) {
        a = tec[i].first;
        while (a--) {
          cout << "col " << tec[i].second << endl;
        }
      }
    } else {
      cout << sum2 << endl;
      for (int i = 0; i < rec.size(); i++) {
        a = rec[i].first;
        while (a--) {
          cout << "col " << rec[i].second << endl;
        }
      }
      for (int i = 0; i < cec.size(); i++) {
        a = cec[i].first;
        while (a--) {
          cout << "row " << cec[i].second << endl;
        }
      }
    }
  } else
    cout << -1 << endl;
  return 0;
}
