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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

//package com.company;

import java.io.*;
import java.util.*;

public class Main {
    static long TIME_START, TIME_END;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("Test.in"));
        PrintWriter pw = new PrintWriter(System.out);
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.out"));

        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc, pw);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
        pw.close();
        System.out.println("Memory increased:" + (usedMemoryAfter - usedMemoryBefore) / 1000000);
        System.out.println("Time used: " + (TIME_END - TIME_START) + ".");
    }

    public static class Task {
        int[] dirx = new int[]{1, -1, 0, 0};
        int[] diry = new int[]{0, 0, 1, -1};
        public int dfs(int[][] grid, boolean[][] visit, int curx, int cury, int fromx, int fromy, List<int[]> op) {
            visit[curx][cury] = true;
            for (int i = 0; i < 4; i++) {
                int nextx = curx + dirx[i];
                int nexty = cury + diry[i];
                if (!visit[nextx][nexty] && grid[nextx][nexty] > 0) {
                    dfs(grid, visit, nextx, nexty, curx, cury, op);
                }
            }
            if (fromx != -1) {
                if (grid[curx][cury] % 2 != 0) {
                    op.add(new int[]{curx, cury, fromx, fromy});
                    grid[fromx][fromy]++;
                    grid[curx][cury]--;
                }
            }
            return grid[curx][cury] % 2 == 0 ? 1 : 0;
        }

        public void solve(Scanner sc, PrintWriter pw) throws IOException {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int[][] grid = new int[n + 2][m + 2];
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                    grid[i][j] = sc.nextInt();
                }
            }
            boolean[][] visit = new boolean[n + 2][m + 2];
            List<int[]> op = new ArrayList<>();
            for (int i = 0; i < n + 2; i++) {
                for (int j = 0; j < m + 2; j++) {
                    if (!visit[i][j] && grid[i][j] > 0) {
                        dfs(grid, visit, i, j, -1, -1, op);

                    }
                }
            }
            pw.println(op.size());
            for (int i = 0; i < op.size(); i++) {
                for (int j = 0; j < 4; j++) {
                    pw.print(op.get(i)[j] + " ");
                }
                pw.println();
            }
        }
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader s) throws FileNotFoundException {
            br = new BufferedReader(s);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}