Both Twins Traveling Paradox

Two twins T$_{1}$ and T$_{2}$ synchronize their clocks at the same location L, then both of them leave with the same uniform high speed $v$ and on the same large distance $d$ on opposite linear directions to the locations $A$ and respectively $B$ (of course \textit{LA = LB = d}) on that planet. Each twin sees the other twin moving away from him with the relativistic speed $2v$, so each twin considers the other twin younger than him. The time dilation is the same in both twins' inertial reference frames. Here it is a forth symmetry. They stop there at $A$ and respectively at $B$. Afterwards, the twin $T_{1}$ from $A$ travels on a linear route back to $B$ (passing through $L)$ at a uniform high speed $2v$. Again, each twin sees the other twin traveling towards him with a speed $2v$. And again each twin considers the other twin being younger than him, since there is the same time dilation and same space contraction. Again one has a back symmetry. But, when the twin $T_{1}$ from $A$ gets to $B$, he finds out that he is younger than the twin $T_{2}$ in $B$ since he has traveled more that $T_{2}$.