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;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC optimize(3)
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC target("sse3", "sse2", "sse")
#pragma GCC target("avx", "sse4", "sse4.1", "sse4.2", "ssse3")
#pragma GCC target("f16c")
#pragma GCC optimize("inline", "fast-math", "unroll-loops", \
                     "no-stack-protector")
#pragma GCC push_options
#pragma GCC pop_options
#pragma GCC reset_options
#pragma GCC optimize("O3")
const double eps = 1e-9;
const int INFMEM = 63;
const int INF = 1061109567;
const long long LINF = 4557430888798830399LL;
const double DINF = numeric_limits<double>::infinity();
const long long MOD = 1000000007;
const int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
const double PI = 3.141592653589793;
inline void fastll(long long &input_number) {
  input_number = 0;
  int ch = getchar_unlocked();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar_unlocked();
  }
  while (ch >= '0' && ch <= '9') {
    input_number = (input_number << 3) + (input_number << 1) + ch - '0';
    ch = getchar_unlocked();
  }
  input_number *= sign;
}
inline void open(string a) {
  freopen((a + ".in").c_str(), "r", stdin);
  freopen((a + ".out").c_str(), "w", stdout);
}
inline void fasterios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long r, c;
bool isi[55][55];
bool ans[55][55];
vector<pair<long long, long long> > allans;
int main() {
  cin >> r >> c;
  for (int i = 1; i <= r; i++) {
    for (int j = 1; j <= c; j++) {
      cin >> isi[i][j];
    }
  }
  for (int i = 1; i < r; i++) {
    for (int j = 1; j < c; j++) {
      if (isi[i][j] && isi[i + 1][j] && isi[i][j + 1] && isi[i + 1][j + 1]) {
        ans[i][j] = ans[i + 1][j] = ans[i][j + 1] = ans[i + 1][j + 1] = 1;
        allans.push_back({i, j});
      }
    }
  }
  for (int i = 1; i <= r; i++) {
    for (int j = 1; j <= c; j++) {
      if (isi[i][j] != ans[i][j]) {
        cout << -1 << '\n';
        return 0;
      }
    }
  }
  cout << allans.size() << '\n';
  for (int i = 0; i < allans.size(); i++) {
    cout << allans[i].first << " " << allans[i].second << '\n';
  }
  return 0;
}
