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 (GCC 5.4.1) (judge verdict: Runtime Error)

/*

結果:
*/
#define _USE_MATH_DEFINES
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <limits.h>  
#include <stdbool.h>   
#define inf (INT_MAX-1)
#define INF 9223372036854775807
#define PI 3.14159265358979323846;
#define EPS 1e-10
#define sq(n) ((n)*(n))
#define rep(i,n) for(i=0;i<n;i++)
#define rev(i,n) for(i=n-1;i>=0;i--)
#define sort(a,n) qsort(a,n,sizeof(TYPE),cmp)
#define sort_r(a,n) qsort(a,n,sizeof(TYPE),cmp_r)
#define chsort(s,n) qsort(s,n,sizeof(char),cmp)
#define chsort_r(s,n) qsort(s,n,sizeof(char),char_cmp_r)
#define TYPE long long
#define ll long long
#define MEMSET(a) memset(a,0,sizeof(a))
#define MEMSET_U(a) memset(a,-1,sizeof(a))
const int mod = (int)1e09 + 7;

//#define DEBUG1
//#define DEBUG2
//#define DEBUGF
#define DUMMY

int in(void) { int i; scanf("%d", &i); return i; }
long long llin(void) { long long i; scanf("%lld", &i); return i; }
double din(void) { double i; scanf("%lf", &i); return i; }
void chin(char s[]) { scanf("%s", s); }

void print(int a) { printf("%d\n", a); }
void llprint(long long a) { printf("%lld\n", a); }
void dprint(double a) { printf("%.10f\n", a); }
void print2(int a, int b) { printf("%d %d\n", a, b); }

int MAX(int a, int b) { if (a > b) { return a; }return b; }
/*
int min(int a, int b) { if (a<b) { return a; }return b; }
*/
long long llmax(long long a, long long b) { return a > b ? a : b; }
long long llmin(long long a, long long b) { return a < b ? a : b; }
double dmax(double a, double b) { return a > b ? a : b; }
int cmp(const void *a, const void *b) { return *(TYPE *)a - *(TYPE *)b; }
int cmp_r(const void *a, const void *b) { return *(TYPE *)b - *(TYPE *)a; }
int char_cmp(const void *a, const void *b) { return strcmp((char *)a, (char *)b); }
int char_cmp_r(const void *a, const void *b) { return strcmp((char *)b, (char *)a); }
void swap(int *a, int *b) { int t = *a; *a = *b; *b = t; }

//A:AC
//int main() {
//	int A = in();
//	int B = in();
//	int c;
//	int d;
//	int flg = 0;
//
//	for (c = 1; c <= 3; c++) {
//		d = (A*B*c) % 2;
//		if (d == 1) { flg = 1; }
//	}
//	if (flg) {
//		printf("%s\n", "Yes");
//	}
//	else {
//		printf("%s\n", "No");
//	}
//
//
//
//#ifdef DEBUGF
//	getch();
//#endif
//	return 0;
//	}

//B:AC
//int main() {
//
//	char w[100][11];
//	int n, i, j;
//	scanf("%d", &n);
//	for (i = 0; i<n; ++i)scanf("%s", w[i]);
//	for (i = 0; i<n - 1; ++i) {
//		for (j = i + 1; j<n; ++j) {
//			if (!strcmp(w[i], w[j])) {
//				puts("No");
//				return 0;
//			}
//		}
//	}
//	for (i = 1; i<n; ++i) {
//		if (w[i - 1][strlen(w[i - 1]) - 1] != w[i][0]) {
//			puts("No");
//			return 0;
//		}
//	}
//	puts("Yes");
//
//#ifdef DEBUGF
//	getch();
//#endif
//	return 0;
//	}

//C:AC
///// /////////////////////////////////////////////
///// <summary> 
///// ユークリッドの互除法
/////機　能：ユークリッドの互除法によりa,bの最大公約数を求める
/////戻り値: 最大公約数、(自然数以外が指定されたときは-1)
///// </summary> 
///// /////////////////////////////////////////////
//int euclid(int a, int b)
//{
//	int temp;
//	if (a < b) { temp = a; a = b; b = temp; }
//	if (b < 1) return -1;
//	if (a % b == 0) return b;
//	return euclid(b, a % b);
//}
//
//int main() {
//	int N, X,A[100000];
//	int a, b,ans;
//	scanf("%d%d", &N, &X);
//	for (int i = 0; i < N; i++) {
//		scanf("%d", &A[i]);
//	}
//	a = abs(X - A[0]);
//	if (N > 1) {
//		for (int i = 0; i < N-1; i++) {
//			b = abs(X - A[i+1]);
//			a = euclid(a, b);
//		}
//	}
//	ans = a;
//	print(a);
//
//#ifdef DEBUGF
//	getch();
//#endif
//	return 0;
//}

//D
int A[500][500] = { 0 };
int ans[2500][4] = { 0 };

int main() {
	int H, W;
	int n = 0;
	scanf("%d%d", &H, &W);
	for (int h = 0; h < H; h++) {
		for (int w = 0; w < W; w++) {
			scanf("%d", &A[h][w]);
		}
	}

	for (int h = 0; h < H; h++) {
		for (int w = 0; w < W - 1; w++) {
			if (A[h][w] % 2) {
				A[h][w]--;
				A[h][w + 1]++;

				n++;
				ans[n][0] = h;
				ans[n][1] = w;
				ans[n][2] = h;
				ans[n][3] = w + 1;
			}
		}
	}
	for (int h = 0; h < H - 1; h++) {
		if (A[h][W - 1] % 2) {
			A[h][W - 1]--;
			A[h + 1][W]++;

			n++;
			ans[n][0] = h;
			ans[n][1] = W - 1;
			ans[n][2] = h + 1;
			ans[n][3] = W - 1;
		}
	}

	printf("%d\n", n);
	for (int i = 1; i <= n; i++) {
		printf("%d %d %d %d\n", ans[i][0] + 1, ans[i][1] + 1, ans[i][2] + 1, ans[i][3] + 1);
	}

#ifdef DEBUGF
	getch();
#endif
	return 0;
}