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 PYTHON):

Solution:
#!/usr/bin/python
# -*- coding: utf-8 -*-

s = raw_input()
H, W = [int(x) for x in s.split()]
# H, W = 500,500
A = [None] * H

for i in xrange(H):
	s = raw_input()
	A[i] = [int(x) for x in s.split()]
	# A[i] = [j for j in xrange(W)]

choose = [[False]*W for i in xrange(H)]
answer = -1
answerOps = []
ops = []

def dfs(y, x, even):
	global H, W, A, answer, answerOps, ops

	if x >= W:
		x, y = 0, y+1
	if x == 0 and y == H:
		if even > answer:
			answer = even
			answerOps = ops
		return
	if even + (H-y-1)*W + (W-x) <= answer:
		return

	# print y, x, A[y][x], even
	# print A

	coin = A[y][x]
	if coin % 2 == 1:
		A[y][x] = 0
		if x+1 < W:
			A[y][x+1] += coin
			ops.append((y, x, y, x+1))
			dfs(y, x+1, even+1)
			ops = ops[:-1]
			A[y][x+1] -= coin

		if y+1 < H:
			A[y+1][x] += coin
			ops.append((y, x, y+1, x))
			dfs(y, x+1, even+1)
			ops = ops[:-1]
			A[y+1][x] -= coin
		A[y][x] = coin

		dfs(y, x+1, even)
	else:
		dfs(y, x+1, even+1)


# dfs(0, 0, 0)


def check(y, x):
	if x+1 < W:
		if A[y][x+1] % 2 == 1:
			return 'right'
	if y+1 < H:
		if A[y+1][x] % 2 == 1:
			return 'down'
	if x+1 < W:
		return 'right'
	if y+1 < H:
		return 'down'

for i in xrange(H):
	for j in range(W):
		if A[i][j] % 2 == 1:
			ret = check(i, j)
			if ret == 'right':
				A[i][j+1] += A[i][j]
				answerOps.append((i, j, i, j+1))
			elif ret == 'down':
				A[i+1][j] += A[i][j]
				answerOps.append((i, j, i+1, j))

print len(answerOps)
for t in answerOps:
	print '%d %d %d %d' % tuple([x+1 for x in t])