Problem: There is a grid of square cells with H horizontal rows and W vertical columns. The cell at the i-th row and the j-th column will be denoted as Cell (i, j).

In Cell (i, j), a_{ij} coins are placed.

You can perform the following operation any number of times:

Operation: Choose a cell that was not chosen before and contains one or more coins, then move one of those coins to a vertically or horizontally adjacent cell.

Maximize the number of cells containing an even number of coins.

Constraints

* All values in input are integers.
* 1 \leq H, W \leq 500
* 0 \leq a_{ij} \leq 9

Input

Input is given from Standard Input in the following format:


H W
a_{11} a_{12} ... a_{1W}
a_{21} a_{22} ... a_{2W}
:
a_{H1} a_{H2} ... a_{HW}


Output

Print a sequence of operations that maximizes the number of cells containing an even number of coins, in the following format:


N
y_1 x_1 y_1' x_1'
y_2 x_2 y_2' x_2'
:
y_N x_N y_N' x_N'


That is, in the first line, print an integer N between 0 and H \times W (inclusive), representing the number of operations.

In the (i+1)-th line (1 \leq i \leq N), print four integers y_i, x_i, y_i' and x_i' (1 \leq y_i, y_i' \leq H and 1 \leq x_i, x_i' \leq W), representing the i-th operation. These four integers represents the operation of moving one of the coins placed in Cell (y_i, x_i) to a vertically or horizontally adjacent cell, (y_i', x_i').

Note that if the specified operation violates the specification in the problem statement or the output format is invalid, it will result in Wrong Answer.

Examples

Input

2 3
1 2 3
0 1 1


Output

3
2 2 2 3
1 1 1 2
1 3 1 2


Input

3 2
1 0
2 1
1 0


Output

3
1 1 1 2
1 2 2 2
3 1 3 2


Input

1 5
9 9 9 9 9


Output

2
1 1 1 2
1 3 1 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

int main(){
  int H,W;
  cin >> H >> W;
  vector<int> w(H*W),m(2*H*W);

  int i,j,k=0;
  for(i=0; i<H*W; i++){
    cin >> w.at(i);
  }

  for(i=0; i<H; i++){
    for(j=0; j<W-1; j++){
      if(w.at(i*W+j)%2 == 1){
        w.at(i*W+j)-=1;
        w.at(i*W+j+1)+=1;
        m.at(2*k)=i*W+j;
        m.at(2*k+1)=i*W+j+1;
        k++;
      }
    }
  }

  for(i=0; i<H-1; i++){
    if(w.at((i+1)*W-1)%2 == 1){
        w.at((i+1)*W-1)-=1;
        w.at((i+2)*W-1)+=1;
        m.at(2*k)=(i+1)*W-1;
        m.at(2*k+1)=(i+2)*W-1;
        k++;
    }
  }

  cout << k << endl;
  for(i=0; i<k; i++){
    cout << m.at(2*i)/W+1 << " " << m.at(2*i)%W+1 << " " << m.at(2*i+1)/W+1 << " " << m.at(2*i+1)%W+1 << endl;
  }
}