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} 
#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long int rand(long long int x, long long int y) {
  ++y;
  return (rng() % (y - x)) + x;
}
string to_string(char c) {
  string second(1, c);
  return second;
}
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void degug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void degug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  degug_out(T...);
}
inline long long int gcd(long long int a, long long int b) {
  if (a > b) swap(a, b);
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long int n, m;
bool grid[52][52], grid2[52][52];
bool doesnt(long long int x, long long int y) {
  if (x + 1 > n || y + 1 > m) return 0;
  return (grid[x][y] && grid[x + 1][y] && grid[x][y + 1] && grid[x + 1][y + 1]);
}
vector<pair<long long int, long long int> > moves;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long int i = (1); i <= (long long int)(n); ++i)
    for (long long int j = (1); j <= (long long int)(m); ++j) cin >> grid[i][j];
  for (long long int i = (1); i <= (long long int)(n); ++i)
    for (long long int j = (1); j <= (long long int)(m); ++j)
      if (doesnt(i, j)) {
        grid2[i][j] = grid2[i + 1][j] = grid2[i][j + 1] = grid2[i + 1][j + 1] =
            1;
        moves.emplace_back(i, j);
      }
  for (long long int i = (1); i <= (long long int)(n); ++i)
    for (long long int j = (1); j <= (long long int)(m); ++j)
      if (grid[i][j] ^ grid2[i][j]) {
        cout << "-1\n";
        return 0;
      }
  cout << moves.size() << '\n';
  for (auto i : moves) cout << i.first << ' ' << i.second << '\n';
}
