dimension than the expected dimension. This means that $C_4$ only deforms in a divisor (it cannot deform in a surface otherwise we can choose the curves $C_1, \ldots, C_3$ to miss it). Then the bend-and-break argument shows that there is a rational curve $D$ of $-K_X$ degree $1$ and passing through $2$ general points in the divisor. It can connect two of the three free curves, say, $C_1$ and $C_2$. Now take the reducible curve $C_1\cup D \cup C_2$. Then the Gromov-Witten invariant $\langle [pt], [pt]\rangle^X_{0, [C_1+C_2+D]}$ is non-zero. In fact, an irreducible curve in this curve class meeting two general points are very free and the moduli space is smooth of expected dimension at this point. And the only possible reducible curve in this class that can pass through $2$ general points is either of the form $C\cup C'$, where $C$ and $C'$ are free curves, or $C+C'+D$, where $C$ and $C'$ are free curves and $-K_X \cdot D=1$. The first case is impossible if we choose the points to be general since every irreducible component of the moduli space has expected dimension at this point. In the second case $D$ only deforms in a divisor ($D$ cannot deform 