Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

2 3
1 2 3
0 1 1

Sample Output 1

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

Every cell contains an even number of coins after the following sequence of operations:

Move the coin in Cell (2, 2) to Cell (2, 3).

Move the coin in Cell (1, 1) to Cell (1, 2).

Move one of the coins in Cell (1, 3) to Cell (1, 2).

Sample Input 2

3 2
1 0
2 1
1 0

Sample Output 2

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

Sample Input 3

1 5
9 9 9 9 9

Sample Output 3

2
1 1 1 2
1 3 1 4

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<algorithm>
#include<numeric>

using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;

class MYCP {
public:

	//数値を区切って文字列にする
	static string MakeString_LongLong(vector<long long> const& numbers,string const& str) {
		if (numbers.size() == 0)return "";
		string result = "" + to_string(numbers[0]);
		for (long long i = 1;i < numbers.size();i++) {
			result += str;
			result += to_string(numbers[i]);
		}

		return result;
	}

	//空白で区切る為のオーバーロード
	static string MakeString_LongLong(vector<long long> const& numbers) {
		if (numbers.size() == 0)return "";
		string result = "" + to_string(numbers[0]);
		for (long long i = 1; i < numbers.size(); i++) {
			result += " ";
			result += to_string(numbers[i]);
		}

		return result;
	}



	//文字列の配列を改行を挟んでまとめる
	static string MakeString_VectorString(vector<string> const& str) {
		string result = "";
		for (long long i = 0; i < str.size(); i++) {
			result += str[i] + "\n";
		}

		return result;
	}

	//文字列を必要な個数だけ読み取る
	static vector<string> MyReadLineSplit(LL n) {
		vector<string> str(n);
		for (long long i = 0;i < n;i++) {
			std::cin >> str[i];
		}

		return str;
	}

	//数値を必要な個数だけ読み取る
	static vector<long long> ReadInts(long long number) {
		vector<long long> a(number);
		for (int i = 0;i < number;i++) {
			std::cin >> a[i];
		}

		return a;
	}

	//渡された自然数が素数ならtureを返す
	static bool PrimeCheck_Int(long long number) {
		if (number < 2)return false;

		for (ULL i = 2;i*i <= number;i++) {
			if (number%i == 0)return false;
		}

		return true;
	}

	//渡された数値以下の素数表を作る
	static vector<long long> MakePrimeList(long long n) {
		vector<long long> list;
		LL i, j, p;
		bool flag;
		for (i = 2; i <= n; i++) {
			flag = true;
			for (j = 0; j < list.size(); j++) {

				if (!(list[j] * list[j] <= i))break;

				if (i%list[j] == 0) {
					flag = false;
					break;
				}
			}
			if (flag)list.push_back(i);
		}

		return list;
	}

	

	//文字列の分割
	static vector<string> split(string const& str, char sep)
	{
		vector<std::string> v;        // 分割結果を格納するベクター
		auto first = str.begin();              // テキストの最初を指すイテレータ
		while (first != str.end()) {         // テキストが残っている間ループ
			auto last = first;                      // 分割文字列末尾へのイテレータ
			while (last != str.end() && *last != sep)       // 末尾 or セパレータ文字まで進める
				last++;
			v.push_back(string(first, last));       // 分割文字を出力
			if (last != str.end())
				last++;
			first = last;          // 次の処理のためにイテレータを設定
		}
		return v;
	}

	//合計を求める
	static LL Sum(vector<LL> a) {
		LL i, sum = 0;

		for (i = 0; i < a.size(); i++) {
			sum += a[i];
		}

		return sum;
	}

	//小文字ならtrueを返す
	static bool Komoji(char a) {
		if (a >= 'a'&&a <= 'z')return true;

		return false;
	}

	//大文字ならtrueを返す
	static bool Oomoji(char a) {
		if (a >= 'A'&&a <= 'Z')return true;

		return false;
	}

	//最初に呼び出す
	static void SetUp() {
		cin.tie(0);
		ios::sync_with_stdio(false);
	}
};

//累積和を求めるクラス
class Syakutori {
private:
	vector<LL> list;

public:

	void MakeArray(vector<LL> data) {
		LL i;
		list = data;
		list.push_back(0);
		list[0] = 0;
		for (i = 1; i < list.size(); i++) {
			list[i] = list[i - 1] + data[i - 1];
		}
		
	}

	LL Sum(LL start, LL end) {
		if (end < start) {
			cout << "startがendより大きいです";
			return 0;
		}
		if (start < 0 || end >= list.size()) {
			cout << "範囲が異常";
			return 0;
		}

		return list[end] - list[start];
	}




};

LL maxD(vector<double> a) {
	LL i, result;
	double max;

	result = 0;
	max = a[0];

	for (i = 0; i < a.size(); i++) {
		if (max < a[i]) {
			max = a[i];
			result = i;
		}
	}

	return result;
}

LL minLL(VLL a) {
	LL i, result;
	LL min;

	result = 0;
	min = a[0];

	for (i = 0; i < a.size(); i++) {
		if (min > a[i]) {
			min = a[i];
			result = i;
		}
	}

	return result;
}


int main(void) {
	MYCP::SetUp();
	LL i, j, k, n, s, x, y, h, w, count = 0;
	cin >> h >> w;
	vector<VLL> field, result;
	if (true) {
		VLL temp(w);
		for (i = 0; i < h; i++) {
			VLL temp2 = temp;
			field.push_back(temp2);
		}
	}

	VLL temp(4), reset(4);

	for (i = 0; i < h; i++) {
		for (j = 0; j < w; j++) {
			cin >> field[i][j];
		}
	}

	for (i = 0; i < h; i++) {
		for (j = 0; j < w - 1; j++) {
			if (field[i][j] % 2 == 1) {
				temp = reset;

				field[i][j] -= 1;
				field[i][j + 1] += 1;
				count++;

				temp[0] = i + 1;
				temp[1] = j + 1;
				temp[2] = i + 1;
				temp[3] = j + 2;

				result.push_back(temp);

			}
		}

		if (i < h - 1) {
			j = w - 1;
			if (field[i][j] % 2 == 1) {
				temp = reset;

				field[i][j] -= 1;
				field[i + 1][j] += 1;
				count++;

				temp[0] = i + 1;
				temp[1] = j + 1;
				temp[2] = i + 2;
				temp[3] = j + 1;

				result.push_back(temp);
			}
		}
	}

	cout << count << endl;
	for (i = 0; i < count; i++) {
		cout << result[i][0] << result[i][1] << result[i][2] << result[i][3] << endl;
	}

	return 0;

}