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 <iostream>
using namespace std;

int main() {
	int n, m, g = 0;
	cin >> m >> n;
	int x[m][n], e[4][m * n];
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n; j++) {
			cin >> x[i][j];
		}
	}
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n - 1; j++) {
			if (x[i][j] % 2 == 1) {
				e[0][g] = i + 1;
				e[1][g] = j + 1;
				e[2][g] = i + 1;
				e[3][g] = j + 2;
				g++;
				x[i][j]--;
				x[i][j + 1]++;
			}
		}
		if (x[i][n - 1] % 2 == 1 && i < m - 1) {
			e[0][g] = i + 1;
			e[1][g] = n;
			e[2][g] = i + 2;
			e[3][g] = n;
			g++;
			x[i][n - 1]--;
			x[i + 1][n - 1]++;
		}
	}
	cout << g << endl;
	for (int i = 0; i < g; i++)cout << e[0][i] << " " << e[1][i] << " " << e[2][i] << " " << e[3][i] << endl;

	return 0;
}