If the quadratic equation $x^2+(2-\tan \theta )x-(1+\tan \theta )=0$ has two integral roots, Then sum of all possible values of $\theta$ in interval $(0,2\pi )$ is $k\pi .$ The value of k is________

 
$\alpha +\beta =\tan \theta -2$ ………………….(1)
$\alpha \beta =-\tan \theta -1$ …………………..(2)
(1) + (2) $\Rightarrow \alpha +\beta +\alpha \beta =-3\Rightarrow (\alpha +1)(\beta +1)=-2$  
Hence either $\alpha +1=-2$ and $\beta +1=1$ or $\alpha +1=-1$ and $\beta +1=2$ 
i.e., either $\alpha =-3$ and $\beta =0$ or $\alpha =-2$ and $\beta =1$   
if $\alpha =-3$ and $\beta =0\Rightarrow \tan \theta =-1\Rightarrow \theta =\frac{3\pi }{4},\frac{7\pi }{4}$  
if $\alpha =-2$ and $\beta =1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4},\frac{5\pi }{4}$  
Hence sum of all possible values of $\theta$ is $0\frac{\pi }{4}+\frac{3\pi }{4}+\frac{5\pi }{4}+\frac{7\pi }{4}=\frac{16\pi }{4}=4\pi .$
Hence k = 4