Dilation equation of motion (2D Dilation gravity)

In the paper "Les Houches Lectures on Black Holes", we have the action given as: (Eqn 3.6),
$$ S = int d^2x sqrt{-g} e^{-2 phi } [R + 4(nabla phi)^2 + 4 lambda^2] $$

This is a 1 + 1-dimensional theory of gravity coupled to
a dilaton field $phi$. The equation of motion for dilation field $phi$ is obtained as Equation 3.8:
$$ R – 4(nabla phi)^2 + 4 lambda^2 + 4Box phi = 0 $$

But, when I try to rederive the equation of motion for the dilation field $phi$, I am getting a different sign.

$$ S = int d^2x sqrt{-g} e^{-2 phi } [R + 4(nabla phi)^2 + 4 lambda^2] $$
$$ delta S = int d^2x sqrt{-g} delta (e^{-2 phi }) [R + 4(nabla phi)^2 + 4 lambda^2] +
e^{-2 phi } delta[R + 4g^{uv}nabla_u phi nabla_v phi + 4 lambda^2]$$
$$ = int d^2x sqrt{-g}(-2e^{-2 phi }) [R + 4(nabla phi)^2 + 4 lambda^2]deltaphi +
e^{-2 phi } (4*2 g^{uv}nabla_u deltaphi nabla_v phi)$$
$$ = int d^2x sqrt{-g}(-2e^{-2 phi }) [R + 4(nabla phi)^2 + 4 lambda^2]deltaphi –
e^{-2 phi } (8g^{uv}nabla_u nabla_v phideltaphi) + text{bdr. terms}$$

Dilation Equation of motion is then:
$$ R + 4(nabla phi)^2 + 4 lambda^2 + 4g^{uv}nabla_u nabla_v phi = 0 $$
$$ R + 4(nabla phi)^2 + 4 lambda^2 + 4Box phi = 0 $$

As you can see the sign in $ 4(nabla phi)^2$ is different than obtained in the paper.