Real numbers $x,y,z$ satisfy $x+xy+xyz=1,y+yz+xyz=2,z+xz+xyz=4.$ The largest possible value of $xyz$ is $\frac{a+b\sqrt{c}}{d},$ where a,b,c,d are integers, d is positive , c is square-free, and $\gcd (a,b,d)=1.$ Then $1000a+100b+10c+d.$