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.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ

		Scanner sc1 = new Scanner(System.in);

		int h = sc1.nextInt();
		int w = sc1.nextInt();
		int count = 0;
		int countA = 0;
		int tmpA = 0;
		int tmpB = 0;
		int tmpC = 0;
		int tmpD = 0;
		int[][] map = new int[h][w];
		String array[] = new String[h*w];

		for ( int i = 0 ; i < h ; i++ ) {
			for ( int j = 0 ; j < w ; j++) {
				map[i][j] = sc1.nextInt();
			}
		}

		for ( int i = 0 ; i < h ; i++ ) {
			for ( int j = 0 ; j < w ; j++) {
				if (map[i][j] % 2 != 0) {
					//下チェック
					if (j < w-1) {
						if (map[i][j+1] % 2 != 0) {
							map[i][j]--;
							map[i][j+1]++;
							count++;
							tmpA = i+1;
							tmpB = j+1;
							tmpC = i+1;
							tmpD = j+2;
							array[countA] = tmpA + " " + tmpB + " " + tmpC + " " + tmpD;
							countA++;
							continue;
						}
					}

					//右チェック
					if (i < h-1) {
						map[i][j]--;
						map[i+1][j]++;
						count++;
						tmpA = i+1;
						tmpB = j+1;
						tmpC = i+2;
						tmpD = j+1;
						array[countA] = tmpA + " " + tmpB + " " + tmpC + " " + tmpD;
						countA++;
					}
					else if (j < w-1) {
						map[i][j]--;
						map[i][j+1]++;
						count++;
						tmpA = i+1;
						tmpB = j+1;
						tmpC = i+1;
						tmpD = j+2;
						array[countA] = tmpA + " " + tmpB + " " + tmpC + " " + tmpD;
						countA++;
					}
				}
			}
		}

		System.out.println(count);

		for ( int i = 0 ; i < countA ; i++ ) {
			System.out.println(array[i]);
		}

		//System.out.println();

		//for ( int i = 0 ; i < h ; i++ ) {
		//	for ( int j = 0 ; j < w ; j++) {
		//		System.out.println(map[i][j]);
		//	}
		//}

		sc1.close();

	}

}