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 JAVA):

Solution:
import java.util.*;

public class Main {

	public static void main(
			String[] args) {
		Scanner sc = new Scanner(System.in);
		int h = sc.nextInt();
		int w = sc.nextInt();
		boolean[][] map = new boolean[h][w];
		for (int i = 0; i < h; i++) {
			for (int j = 0; j < w; j++) {
				map[i][j] = sc.nextInt() % 2 == 1;
			}
		}
		ArrayList list = new ArrayList();
		// 一筆書き
		int cnt = 0;
		while (cnt < h * w) {
			int[] now = getPosition(cnt, w);
			if (map[now[1] - 1][now[0] - 1]) {
				int[] next = getPosition(cnt + 1, w);
				if (next[1] > h) {
					break;
				}
				list.add(now[1] + " " + now[0] + " " + next[1] + " " + next[0]);
				map[now[1] - 1][now[0] - 1] = !map[now[1] - 1][now[0] - 1];
				map[next[1] - 1][next[0] - 1] = !map[next[1] - 1][next[0] - 1];
			}
			cnt++;
		}
		System.out.println(list.size());
		for (int i = 0; i < list.size(); i++) {
			System.out.println(list.get(i));
		}
	}

	static int[] getPosition(
			int cnt,
			int w) {
		int y = cnt / w + 1;
		int x = (y % 2 == 0 ? w - cnt % w : cnt % w + 1);
		int[] xy = {
			x,
			y };
		return xy;
	}
}