#include <bits/stdc++.h>
using namespace std;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 5e1 + 7;
int n, m, a[N][N], b[N][N];
vector<pair<int, int> > v;
bool check(int a, int b, int c, int d) {
  return a == 1 && b == 1 && c == 1 && d == 1;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= m - 1; j++) cin >> a[i][j];
  for (int i = 0; i <= n - 2; i++)
    for (int j = 0; j <= m - 2; j++) {
      if (check(a[i][j], a[i + 1][j], a[i][j + 1], a[i + 1][j + 1])) {
        b[i][j] = b[i + 1][j] = b[i][j + 1] = b[i + 1][j + 1] = 1;
        v.push_back(make_pair(i + 1, j + 1));
      }
    }
  bool valid = true;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= m - 1; j++)
      if (a[i][j] != b[i][j]) valid = false;
  if (valid) {
    cout << v.size() << '\n';
    for (auto i : v) cout << i.first << " " << i.second << '\n';
  } else {
    cout << -1 << '\n';
  }
  return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0.

You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2 × 2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1 ≤ x < n and 1 ≤ y < m, and then set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1.

Your goal is to make matrix B equal to matrix A. Two matrices A and B are equal if and only if every element of matrix A is equal to the corresponding element of matrix B.

Is it possible to make these matrices equal? If it is, you have to come up with a sequence of operations that makes B equal to A. Note that you don't have to minimize the number of operations.

Input

The first line contains two integers n and m (2 ≤ n, m ≤ 50).

Then n lines follow, each containing m integers. The j-th integer in the i-th line is A_{i, j}. Each integer is either 0 or 1.

Output

If it is impossible to make B equal to A, print one integer -1.

Otherwise, print any sequence of operations that transforms B into A in the following format: the first line should contain one integer k — the number of operations, and then k lines should follow, each line containing two integers x and y for the corresponding operation (set B_{x, y}, B_{x, y + 1}, B_{x + 1, y} and B_{x + 1, y + 1} to 1). The condition 0 ≤ k ≤ 2500 should hold.

Examples

Input


3 3
1 1 1
1 1 1
0 1 1


Output


3
1 1
1 2
2 2


Input


3 3
1 0 1
1 0 1
0 0 0


Output


-1


Input


3 2
0 0
0 0
0 0


Output


0

Note

The sequence of operations in the first example:

\begin{matrix} 0 & 0 & 0 & & 1 & 1 & 0 & & 1 & 1 & 1 & & 1 & 1 & 1 \\\ 0 & 0 & 0 & → & 1 & 1 & 0 & → & 1 & 1 & 1 & → & 1 & 1 & 1 \\\ 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 0 & 0 & & 0 & 1 & 1 \end{matrix} 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.st << "," << P.nd << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 5e1 + 7;
int n, m, a[N][N], b[N][N];
vector<pair<int, int> > v;
bool check(int a, int b, int c, int d) {
  return a == 1 && b == 1 && c == 1 && d == 1;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= m - 1; j++) cin >> a[i][j];
  for (int i = 0; i <= n - 2; i++)
    for (int j = 0; j <= m - 2; j++) {
      if (check(a[i][j], a[i + 1][j], a[i][j + 1], a[i + 1][j + 1])) {
        b[i][j] = b[i + 1][j] = b[i][j + 1] = b[i + 1][j + 1] = 1;
        v.push_back(make_pair(i + 1, j + 1));
      }
    }
  bool valid = true;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= m - 1; j++)
      if (a[i][j] != b[i][j]) valid = false;
  if (valid) {
    cout << v.size() << '\n';
    for (auto i : v) cout << i.first << " " << i.second << '\n';
  } else {
    cout << -1 << '\n';
  }
  return 0;
}

```