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.*;
import static java.lang.System.*;
import static java.lang.Math.*;
public class Main {
public static void main(String[] args){
Scanner sc = new Scanner(in);
//処理
int H=sc.nextInt();
int W=sc.nextInt();
int[][] a=new int[H][W];
for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
        a[i][j]=sc.nextInt();
    }
}
String[] s=new String[W*H*2];
int c=0;
for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
        if(a[i][j]%2!=0){
            if(i!=H-1){
                s[c+1]=String.valueOf(i+1)+" "+String.valueOf(j+1)+" "+String.valueOf(i+2)+" "+String.valueOf(j+1);
                c++;
                a[i][j]--;
                a[i+1][j]++;
            }else if(j!=W-1){
                s[c+1]=String.valueOf(i+1)+" "+String.valueOf(j+1)+" "+String.valueOf(i+1)+" "+String.valueOf(j+2);
                c++;
                a[i][j]--;
                a[i][j+1]++;
            }
        }
    }
}
s[0]=String.valueOf(c);
for (int j = 0; j <= c; j++) {
    out.println(s[j]);
}
}
//階乗
static long fact(long a){
    long b=1;
    for (long i = 1; i <= a; i++) {
        b*=i;
    }
    return b;
}
//最大公約数
static int gcd (int a, int b) {
    int temp;
    while((temp = a%b)!=0) {
        a = b;
        b = temp;
    }
    return b;
}
//※最小公倍数はa*b/gcd(a,b)である
//二項係数
static long combi (long a, long b) {
    b=min(a-b,b);
    return permu(a,b)/fact(b);
}
//順列
static long permu(long a,long b){
    long temp=1;
    for (long i =a ; i > a-b; i--) {
        temp*=i;
    }
    return temp;
}
}